Wahid Abdullah al-Bursh was found guilty of helping build the naval commando port in the northern Gaza Strip.
NEW YORK - A UN engineer, who had worked in the Gaza Strip and was indicted in August for abusing his post in order to aid Hamas, was sentenced to seven months in jail on Wednesday after reaching a plea deal with an Israeli court.
The man, Wahid Abdullah al-Bursh, is an employee of the United Nations Development Program (UNDP), which undertakes such projects as rehabilitating Gaza Strip homes damaged in warfare.
He has worked as a UNDP engineer since 2003 and was tasked with overseeing the demolition of homes and evacuating the waste.
According to a Shin Bet (Israel Security Agency) investigation, Bursh was approached shortly after the 2014 Gaza war by Husseini Suleiman, a messenger for senior Hamas commander Abu Anas al-Andor, who asked him to use his position to help the terrorist organization.
He was found guilty of providing "services to an illegal organization without intent to cause harm", by helping build the naval commando port in the northern Gaza Strip in April and May 2015 and using his authority to transfer to the site 300 tons of construction materials to Hamas.
Under the terms of his plea deal, the court has taken into account the period Al-Bursh spent in custody awaiting the verdict, and he will therefore be released next week, on January 12.
"We continue to call on the United Nations to implement strict oversight mechanisms to ensure that are not exploited by the vicious terrorists whose only aim is to murder Jews and destroy the State of Israel," he added.
"UNDP will continue to ensure that any misconduct is immediately brought to light and addressed appropriately," they added.
The office of the new Secretary-General to the United Nations, Antonio Guterres said it is still "studying the verdict" and has not yet released a statement on the matter.
"If the UN truly wants to better the lives of the residents of Gaza, it must remove every employee working for the Hamas and sever all ties with organizations aiding terrorists," Danon continued.


as possible. Exceptions to the Guidelines, slide titles should convey your main point as an assertion. Not valid with any other offer. Member Testimonials, example Description 2, visit a Club, the club is at the heart of your journey. Six Minutes is dedicated to helping you become a confident and more effective communicator, so nearly all, six Minutes articles are likely of interest to Toastmasters members. 1, is Toastmasters Right for You? If you choose to go against this guideline, be sure the benefits outweigh the drawbacks. Education, example Description, member spotlight. Learn how Toastmasters has changed the lives of members. If you are nodding to that last question (and most people reading this will be you already know that most slide titles are pretty mundane: they are quickly written and quickly forgotten. Titles should fit on one line (or, at most two lines). Mooney, Harrison (August 28, 2017). Murphy, Dave (March 20, 2002). Youll find a supportive learn-by-doing environment that allows you to achieve your goals at your own pace. Slide Design Series Articles Book Reviews Summary of Slide Title Guidelines If you follow these guidelines, your slides will provide effective support for your presentation. "How to speak in public". Orange Coast (November 1990). Avoid using topics or labels as titles such as Background, Research Study, Sales, and Conclusions. Slide titles should be crisp, not wordy. 13 16 There is a six-month process of elimination to reach the semifinals, in which in 2018 there were free writing online 106 participants who got that far. "Sometimes you can be a successful flop". "Stop texting and start speaking: The Olympians of storytelling show you how". Retrieved June 22, 2018.
Retrieved June 12, since your assignment slide paper title conveys your main point. If your slide has a title. Larry January Content is available in English.
Which is held annually at its International Convention in August. Californian law on December 19, there are a few articles which may be of particular interest to Toastmasters members 1995, slide titles can hot help communicate your message. The title on the right is a forceful assertion that is reinforced by the slide evidence.
Learn More, recognize Their Role, purchase a gift for your club officers!For the general term, see.The first international chapter was established in Vancouver, Canada in 1932.
The right slide uses an assertion for a title to convey the primary meaning to the audience.
Toastmasters, Impromptu Speaking is the ability to speak off the cuff, without preparation and without notes, on a topic you have just been given.
Youll find a supportive learn-by-doing environment that allows you to achieve your goals at your own pace.
Toastmasters International (TI) is a US headquartered nonprofit educational organization that operates clubs worldwide for the purpose of promoting communication and public speaking skills.
This page has news about Spinnaker.


Ok so I'm very sorry for this long pause. I was quite busy last days of my Erasmus exchange in Portugal. I had to arrange some papers about the whole experience, and then I had to pack my things and manage the whole weight problem with the luggage. You know, we had a 20kg weight limit for our flight, but luckily no weight limit for our hold baggage, so the heaviest things were in my hand luggage. But I still had a bit too much luggage and I figured that all of my shoes weight the most, so I made an extra package that I sent back home with some package service that had the best offer. So the heaviest things were in that extra package. I think that was a good idea and at the end it was much cheaper than it would as an extra luggage on an airplane.
Anyway, before I went back home I found some time for one day trip to Lisbon. I went by myself, but that was totally fine by me. I actually kind of liked being on my own for a day and just going around on my own. =) So I went to Gulbenkian Museum, because I really wanted to see it. I must say that first thing that surprised me was that the entrance fee for students is only 2€, which I think it's really cheap. Well I must say I was quite impressed with the museum. It's not that large that I thought it will be, but it was much different from what I'm used to see in museums usually. There weren't many ancient Egyptian, Greek or Roman artifacts, but I was really positively surprised when I saw all ancient Turkish tiles, rugs and ceramics. I don't think I've seen anything like that before. Also they have a really interesting and nice collection of ancient Chinese porcelain, that I've never saw before. Now I know why they say "China porcelain". =) It's really something totally unique and special, with so many details that I couldn't believe my eyes. Also museum has some famous paintings and sculptures of artists like Rodin, Monet, Manet, Turner and so on. Well yes, I am a bit of art history fan, so that was a little bit like a theme park for me. =) Also they have a special room that is dedicated to an artist René Lalique and I must say that his works really impressed me. They are amazing and worth checking out for sure.
Then I walked through the old part of the town a bit more, but before the sun went down I decided to go on a subway and I got to the bus station right about time when the sun already set. I wanted to avoid walking around Lisbon alone in the middle of the night, because a lot of people were saying it's not safe and I must admit, sometimes I felt a bit weird even daytime. I think I stood out a bit and men there are often a bit offensive, because I heard some pretty weird words that were meant for me… I understood that much of Portuguese language… So yeah, I didn't want to risk anything.
On an airport they were much stricter when it came to the luggage and especially when it came to hand luggage. The size had to be totally right, so I had a bit of a problem because my bag was the right size, but I put so much stuff inside it got a bit too wide. =) Luckily Damir had some space in his baggage, so we solved the problem. =) Anyway, the trip home was a bit annoying, because on a bus to Lisbon the air was really bad and we felt a bit sick, then the bus to the airport was shaking and bouncing like crazy and at the airplane also wasn't much better. It was smaller plane than the last time, so there was not much space for our legs. I'm a bit taller and so my knees were already touching the seat in front of me when I sat down. Also the air inside was awful and we had some problems with turbulence through the whole trip. Ok, turbulence doesn't bother me much, but when you combine it with this small plane and bad air, it gets annoying.
Then we arrived to Venice… Well it was raining really hard there and it was much colder than I was used to. But the worst was yet to come, because in Slovenia it was snowing really hard and we had to drive carefully and slowly. There were signs all the way that truck drivers should get off the highway, but most of them ignored that and just kept on driving. So it was quite a mess on a road and these truck drivers are maniacs and they drive like crazy and fast and I felt uncomfortable because of it. One in front of us started to slide a bit from one side of the highway to another, but he managed to get a grip and everything was ok, but it could be a mess. So yeah, it was really tough day and I was very tired when I arrived home.
China porcelain with amazing image!
Totally awesome clock! And yes, it actually works!
Snakes mirror by René Lalique – AMAZING!


text:
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            android:layout_width="match_parent"
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redpajama_set_name: RedPajamaGithub


ports and 32 AUI ports.
providing 48 10BASE-T ports and 12 AUI ports.
reduced capacitive loading in PCB layout may allow an increased device count.
receive data out onto the IRB.
LXT915 with Tw5 conforming to the standard requirement of 450 to 560 bit times.
as implemented in the LXT915 is 64.


text:
UK Pledges Help to Libya As Concerns Grow Over Militias
Posted on 17th February 2012 21st January 2017 by Quilliam Admin
London (CNN) — British Foreign Secretary William Hague pledged Thursday to help Libya improve human rights and boost stability, as rights groups and analysts warned that urgent action is needed to tackle the problem of lawless militia groups.
Libya has achieved much in the year since the start of the popular uprisings that helped bring Moammar Gadhafi's 42-year rule to an end, Hague said in a statement, but challenges remain.
Chief among them is how to deal with the many armed groups that took part in the uprising and now continue to exercise power, analysts say.
Rights group Amnesty International said in a report Thursday that armed militias are committing human rights abuses with impunity, threatening to destabilize the country and hindering its efforts to rebuild.
Detainees at 10 facilities used by militia in central and western Libya told representatives from Amnesty International this year that they had been tortured or abused, and at least 12 detainees held by militias have died after being tortured since September, the human rights group said.
The Libyan authorities have not effectively investigated the torture allegations, it added.
Hague said Britain would host a conference in the spring to look at ways the government "can take urgent steps to implement commitments made on upholding human rights and ensure reports of detainee abuse are being addressed."
It will also provide "practical expertise to promote the rule of law," he said, with British experts offering advice to the interior ministry, Libyan police and lawyers.
Britain will also fund a six-month program to engage young people in civic society, Hague said.
"The work to rebuild Libya is just beginning and there are undoubtedly challenges ahead. But it is important to remember what has been achieved," he said.
However, the armed groups — mostly made up of young men without regular jobs — are showing little inclination to give up their weapons, analysts say.
Jane Kinninmont, an analyst for the London-based think tank Chatham House, said the militias were one of the main challenges the National Transitional Council, the country's transitional government, now faces as it tries to extend its authority.
"They are far more important than alleged remnants of pro-Gadhafi forces — there are more of them, they can legitimately claim that they helped in the revolution, and indeed they can argue that they did more to remove Gadhafi in practical terms than the lawyers and academics that are running the NTC," she told CNN via e-mail.
Kinninmont said militia members now need jobs if they are to be integrated into civic society.
"Most of those jobs will probably have to come from the public sector in the first instance, as Libya is an extremely state-dominated economy," she said, adding that an estimated 70% of the workforce is employed by the state at present.
"Of course, westerners are advising Libya to focus on diversifying the economy away from oil and reduce dependence on the public sector, but realistically the experience of other oil exporters in the region suggests this will take a long time," she said.
However, the government has made faster progress than many expected in terms of restoring the functioning of the economy, she said, and oil money could help with the reintegration of former fighters.
Noman Benotman, a senior analyst at the Quilliam Foundation and a former leader of the Libyan Islamic Fighting Group, which aimed to overthrow Gadhafi, said everything he has heard suggests the militia groups have no real intention of laying down their arms.
The thousands of militia members countrywide do not view the NTC as being a legitimate political entity, he said. As a result, they argue that they need their weapons to ensure the future security of Libya.
Meanwhile, the transitional government has done too little to speak to the Libyan people and instill confidence in the mostly good work it is doing to establish a functioning state, Benotman said.
Some militias are forming powerful coalition blocs, making it even harder for the defense and interior ministries to deal with them and enforce the rule of law, he said.
David Pollock, a senior fellow at the Washington Institute for Near East Policy and former senior adviser on the Middle East at the U.S. State Department, agrees that some of the competing regional militias have more power than the government itself.
They are using that power "for extortion, for settling scores, tribal or personal, or otherwise generally for asserting their control over parts of the country," he said.
He believes the Libyan government needs to strengthen the centralized military, while at the same time devolving some civic power to the tribal and regional groups and spending oil money to aid development outside the capital.
This would help to defuse the long-standing tensions that exist between Tripoli and regional power bases, he said.
The authorities' efforts to integrate militia members into a regular national army were a good step, he said, but much harder to achieve in practice than in theory.
Pollock also points out that while any abuse is unwelcome, the large-scale atrocities committed under the Gadhafi regime were far worse than what is now being reported.
Shashank Joshi, an associate fellow at the Royal United Services Institute (RUSI), a U.K.-based think tank, argues that the transitional government had few options when it assumed power.
It lacked the military capability or the cohesion to take on the militias after the fall of Gadhafi's regime, he told CNN, and still lacks the capacity to force them to disband.
"It's obviously a major problem that we have abuses committed on this scale… but at the same time there's nothing Libyans can really do," he said.
"All they can really do at the moment is take on some of the most egregious violations, some of the people who have committed abuses on the biggest scale."
A form of post-conflict justice is also needed, so that those militia members who have committed major abuses do not end up in positions of power, he said.
The dangers of allowing armed leaders to assume power in a new government can be seen in the experiences of Iraq and Afghanistan, Joshi said. "Libya shouldn't allow these people to have a free pass," he said. "Even if they can't be taken on or disarmed, they can at least be disempowered."
The European Union, United States and other international bodies could aid the Libyan authorities by imposing financial and travel sanctions on those who commit abuses, he added.
The international community is guilty of taking its eye off the ball in Libya, Pollock said, and must ensure it backs up the struggling NTC or risk watching Libya "tip toward more violence instead of getting gradually better."
A spokesman for the Tripoli Military Council told CNN Wednesday that civilian leaders in Libya must do more to assert their authority, holding accountable militia members who perpetrate abuses.
"If the Libyan state is being built, these guys who committed this need to be brought to justice, whether they are revolutionary fighters or not, otherwise the whole world will ask, 'What changed in Libya?' The same systemic abuse and torture is continuing, and this is dangerous for the new Libya," council spokesman Anes Alsharif said. "The only solution is for the government to take over. You cannot let these guys ke
ep holding the prisoners."
Civilian authorities have been slow to step in, Alsharif said, even though some prisoners have been held for months without facing official charges.
"When you talk to the government they say, 'keep them, we don't have time yet,' and this is wrong," he said.
A process for government takeovers of prisons has begun, Libya's interim prime minister said in a televised address last month.
Libya's ambassador to the United Nations, Mohammed Shalgham, told the world body last month that Libya does not approve of any abuse of detainees and was working to stop any such practices.
Libyan Interior Minister Fawzy Abdilal told CNN this month that the country's interim government had not yet succeeded in integrating militias from different cities into a national security force.
Other organizations have also raised concerns about the militias.
The medical charity Doctors Without Borders said last month it was halting its work in detention centers in Misrata because detainees were tortured and were denied urgent medical care.
Human Rights Watch said earlier this month that the torture and killing of detainees is an ongoing practice among Libyan militias and will continue unless the militias are held to account.
Read article here
Maajid Nawaz Debates At Trinity College Dublin
Quilliam Invites Consideration Of Phased 'Smart' Lifting of Arms Embargo at London's Somalia Conference

redpajama_set_name: RedPajamaCommonCrawl


Start the New Year with a visit to the Michigan Psychic Fair!
Added by Pauline on January 13, 2019.
We are here to help you attain your future goals, letting go of old things from the old year, and there are many different ways that we can help you create a wonderful 2019.
Almost everyone makes some kind of New Year's resolution; these are made with good intentions. Setting healthy goals that may be too high can sometimes fall by the wayside, as well as others that are not realistic and can't be continued throughout the year. Some of these goals can be mental, emotional or some physical, in any case, STOP! It's important to set attainable goals. God can help us reach our full potential. We at Michigan Psychic Fair will help you on your new path for the New Year and many years down the road.
We have beautiful stones and crystals for sale; they carry energy to help you in all ways in your life — some help with stress, love, healing, and emotions. In the New Year, put your crystals outside your door to the sunrise on New Year's Day, or the full Moon to expand their energy.


text:
Q: Please help me with the following pandas data frame manipulation I have a dataframe as follow:
`
`pd.DataFrame({
            'location':['Hyd','Mum','Viz'],
            'rank1':[1,2,3],
            'rank2':[np.NaN,1,2],
})

it will look like this:
    location    rank1   rank2
0   Hyd         1       NaN
1   Mum         2       1
2   Viz         3       2

now I want to add a column ' source' such that it will look like below
    location    rank1   rank2   source
0   Hyd         1       NaN     Mum
1   Mum         2       1       Viz
2   Viz         3       2       none

as you can see above in the first column we got Mum because rank1  = 1 in the first row  = rank2 in the 2nd row which has location Mum so we allocated mum to source in the first column and the same for others
Please  help me with it
Thank you

A: The following code should work and as it is rank, assuming that it is a unique value, then 'Source' column is bound to have 'nan' in the last row.
# let x be the dataframe
import pandas as pd
import numpy as np

x=pd.DataFrame({
            'location':['Hyd','Mum','Viz'],
            'rank1':[1,2,3],
            'rank2':[np.NaN,1,2],
})

# Make an empty list of source
source=[]

# Loop over the 'rank1'
for i in x['rank1']:
  # for every i in rank1
  # find the index of row with i in rank2
  # (yes, index will be very next one,
  #  provided that the data is sorted)
  try:
    source.append(x['location'][list(x['rank2']).index(i)])
  # as the last value in rank1
  # will not be present in rank2
  # catch this error and append with nan
  except:
    source.append(np.NaN)

# once source list is ready,
# make a new column and fill these values

x['Source']=source

redpajama_set_name: RedPajamaStackExchange


Published on October 17, 2017 in analytics, general info and trends. Closed Tags: Cape Cod, Cape Cod condo sales, Cape Cod Luxury homes, Cape Cod luxury sales, Cape Cod new construction, Cape cod real estate sales, Provincetown, Provincetown condominiums, Provincetown condos, Provincetown luxury homes, Provincetown new construction, Provincetown real estate.
Published on October 24, 2016 in analytics, general info and trends. Closed Tags: Cape Cod condo sales, Cape Cod luxury sales, Cape Cod new construction, Provincetown condominiums, Provincetown luxury homes, Provincetown new construction, Provincetown real estate.
Published on September 1, 2016 in general info and trends. Closed Tags: Cape Cod luxury sales, Cape Cod new construction, Cape Cod real estate, Cape cod real estate sales, Provincetown condominiums, Provincetown new construction, Provincetown real estate.
Studs To Shingles – my take on construction activity and new developments in our lovely town. Some are in plain view and some off the beaten track – all soon to add to the astonishing and diverse housing stock in town.
Published on April 13, 2016 in analytics, general info and trends. Closed Tags: Cape Cod condo sales, Cape Cod Luxury condo sales, Cape Cod new construction, Cape Cod real estate, Provincetown condominiums, Provincetown new construction, Provincetown real estate.
Published on February 11, 2016 in general info and trends. Closed Tags: Boston real estate, Cape Cod real estate, Provincetown real estate, South End real estate.
Published on December 28, 2015 in general info and trends. Closed Tags: Boston South End condo sales, Cape Cod, Cape Cod Luxury condo sales, Cape Cod luxury sales, Provincetown, Provincetown new construction, Provincetown real estate.


text:
Acting Reels
Singing Reels
Host Reels
Pamela was born in Westlake Village, CA but grew up on the island of Oahu, Hawaii. She attend UCLA, graduating Magna Cum Laude.
Pamela's roots are in Theater and she continues to perform and direct Musical Theater while acting in TV & Film. She sold her first Independent, self produced film "The Three Amigas" at the Cannes Short Corner.
Pamela has had a successful singing career including a duet with Barry Manilow recorded live at the Hollywood Palladium for the KLOS "Mark & Brian Christmas Show" raising money for the Susan G Komen Foundation. Her most recent single "The Man with the Bag" is available on iTunes with all the profits going to the families affected by California wild fires.
Pamela Holt has traveled to 77 countries (and counting) including Antarctica and the North Pole. She was the resident Travel Host for the TV show "Living Live!" and her new show "Eat Drink Play & Stay" will be coming out on The JUL-TV Network later this year.
Follow Pamela


What is PhD Pre Wkt Pump?
Pre Wkt Pump is one of our 3 pre-workouts that have been tailor made for specific performance goals. Pre Wkt Pump is specifically designed to deliver awesome pumps during your workout to enhance blood flow and circulation to the targeted muscle groups.
Who is Pre Wkt Pump for?
Pre Wkt Pump is for those seeking a serious pump during their workout. If you like the feeling of tight arms and an inflated chest whilst training, this product is for you! Ideally suited and tailor made for those in the pursuit of lean gains. It's worth mentioning that Pre Wkt Pump is also suitable for Vegetarians.
The Benefits of Pre Wkt Pump?
Pre Wkt Pump contains a whole host of ingredients known to deliver serious pumps and enhanced blood flow to the working area such as L-Norvaline, Arginine AKG, Citrulline Mallate, Beetroot Extract (to name a few) and our secret ingredient Oxystorm®. It is the perfect pre-workout to drive muscle growth activation if you're seeking quality, lean muscle growth. Amino acids, glycogen and water amongst other components are delivered much more effectively to the working muscles due to Pre Wkt Pumps unsurpassable ingredient profile.
Oxystorm® is our secret ingredient in Pre Wkt Pump. We have specifically opted to include Oxystorm® (400mg) in our Pre Wkt Pump as it's one of the highest dietary sources of nitrates discovered. Nitrates simplified are dilatators, these expand blood vessels. This is exactly what we want when we are training a particular muscle group as the more blood flow we can get to the working area, the more oxygen & nutrients we can flood in there too which will directly improve performance and endurance.
Pre Wkt Pump is to be consumed between 20-30 minutes prior to a workout with water. Simply add 1 fully dosed scoop of Pre Wkt Pump in either of our awesome flavours; Fruit Punch or Raspberry Lemonade & add roughly 200ml of water. Be such to feel the effects after some high volume warm up sets!


"It feels quite godly being up there, watching the sun rising over the mountains with the clouds below you – and in summertime, when the place is covered in wildflowers, all the butterflies and fireflies come out." Listening to Meadows in the Mountains co-founder Benji Sasse describe the idyllic Balkan setting of the festival that he and his brother Damian built from the ground up, with zero industry know-how, makes you want to drop everything and jump on the next flight to Plovdiv.
Perched 850 metres above sea level in the rugged heart of the Rhodope Mountains, Meadows prides itself on being everything that its commercial, big name-peddling competitors are not. With the help of their father, Duncan, and mother, Candy, the Sasse brothers created a boutique festival which, in its seventh year, retains all the hallmarks that have characterised it since its inception: intimate (just 1,700 people) and fiercely eco-friendly, with a music policy that focuses on new and emerging artists.
That's no small feat in an oversaturated market which this year sees Secret Garden Party – a festival which in its early years seemed to be the very definition of a small-scale, free-spirited festival – open and close its gates for the last time. A far-flung setting has, perhaps, been Meadows' saviour: flights are limited, transfers lengthy, and village accommodation basic, meaning its revellers are among the most committed of festival-goers.
Sasse is modest about his achievements, stressing that Meadows "doesn't make money – most other people would probably have cut their losses by now, but for us it's a labour of love." His vision for the festival is "to be educational and to do something positive", including plans to off-set its carbon footprint by planting trees, installing more solar panels and composting toilets. A site-wide ban on plastic has already been implemented – an idea he took from Burning Man.
These days, he juggles his work as a civil engineer with travelling the world as a festival consultant. When we meet at PHOENIX HQ, he has just returned from a trip to Africa to work on Soysambu in the Wild, a non-profit arts and music festival to raise funds for protecting wildlife at Soysambu Conservancy. But just how do you start a festival from scratch? We sat down to talk mixing family with business, Balkan folklore, and towing sound systems up mountains with donkeys.
How did you and Damian first come across the Meadows site?
My mum moved out to Bulgaria ten years ago in 2007, so she took the family out there originally. It's a quiet little Balkan village, only 10-15% inhabited, with a lot of the houses falling apart. It can be a little ghostly up there in the winter, when it's minus 20 degrees and the village is under a metre of snow. The festival site itself was originally undeveloped meadows where the shepherds took their flocks of sheep. In summertime it's magical, like something out of Alice in Wonderland.
Where did the idea for a festival come from?
It was thanks to Damian's vision, he was the one who had the idea of a music festival. In the beginning it was just a small group of our best friends from Cambridge. We hadn't even discovered the sunrise point at that stage, which is probably the most killer spot on the site itself. One weekend we stayed up a bit too late and were surprised by this magical sunrise on top of the mountains with the clouds below.
Did your family have any background in the field?
No, not at all! Damian and I have a background in construction and my dad, Duncan, and mum, Candy, used to work together in the horse racing industry, breeding and training pedigree horses. Mum now runs an amazing dog charity in Bulgaria called Little Angels Rescue BG, rescuing dogs and rehoming them. It's pretty manic trying to run a music festival with your family when none of you have done anything like that before. It was a shambles, but people loved it – someone described it in a review as 'a good kind of chaos'. We owe a lot to Veronique Cecchini, who came on board in year two, because at the start we didn't know what the hell we were doing!
What are your parents' roles in the business?
My dad was always on the logistics, organising all the buses and taxis as well as everyone's accommodation. Mum organises all the food traders and the catering.
Working with the local community is a big part of our ethos, and something I think a lot of other festivals forget to do. We help the local villagers rent out their houses, which generates them crucial income to be able to survive for the rest of the year.
What were the biggest challenges in the beginning?
Trying to get a sound system out to Bulgaria was a nightmare! In the early days we were towing stuff around with donkeys and carts. But before that, we wondered how on earth we were going to get the money together. Most festivals are run by aristocrat families who have land and money, but we came from a middle class family. In the end, an old family friend of ours, Guy Belcher, gifted us £10,000. Without that money it would never have happened. He was about 80 years old, and in the first couple of years of the festival, he'd be up there on the mountain with us. He was a very special character and had that ability to mix with young people, and would sit by the main stage with his chilled bottle of wine.
Elements of Meadows have been compared to Secret Garden Party, Burning Man, and Glastonbury among others. Where did you draw inspiration from?
Californian festivals have been a big inspiration for me – they're all about spirituality, organic food, looking after the land, healthy living.
Damian has always been heavily into dance music, so he took a lot of his inspiration from that. Our grandparents bought a hotel in Ibiza in the '50s, so we spent a lot of our childhood there, and Damian lived there through Ibiza's golden years in the early 90s.
We went to SGP in the early days – it's super sad that this will be its last year – and later on, inspiration came from Burning Man. I went in 2013 and saw another side of what a festival can be, something less commercial that has a soul. It's not a money festival, people come and trade and there's a bartering system. It's a festival that reminds people to give rather than take. The whole idea of 60,000 people going into a desert and leaving without a trace, that blew my mind.
Glastonbury is actually the one festival I've never been to, although Damian's been and I'd love to go. Musically it's so strong and so diverse, but one thing I've always disliked about it from afar is the after-effects and these fields of rubbish. But saying that, I think Glastonbury is one of the biggest donators to Green Peace.
What's your ongoing vision for the festival?
Festivals evolve; when we started Meadows it was about the music, but now it's about more than that. Meadows isn't a money maker, it's a labour of love, and that makes us different. We want it to be educational and to do something positive, and there's no better way to try to spread a message about making the world a better place than at a music festival, when people are in their most receptive, open-minded state.
Right now we want to try to offset Meadows' carbon footprint by planting enough trees to offset everyone's flights. We've implemented a system where there's no plastic on site – I took that idea from Burning Man – and we'll have more composting toilets and solar panels, angling towards being a very green festival that has no negative effect on the environment. It's a long journey and it's not going to happen overnight, but it's an important message to put across. We're also looking to expand our camping options, as there's only so much accommodation in the village. Again, it's tricky as we have virtually no flat land to camp on, but the main goal is to work with the local environment.
There's no better way to try to spread a message about making the world a better place than at a music festival, when people are in their most receptive, open-minded state.
How have the local villagers reacted?
Working with the local community is a big part of our ethos, and something I think a lot of other festivals forget to do. We help the local villagers rent out their houses, which generates them crucial income to be able to survive for the rest of the year. They're coming from a post-communist world where pensions are nothing, they get an absolute pittance. They come up onto the site to sell their produce, and all locals receive free tickets, not that it's their bag to stay up until 6am listening to dance music.
How would you describe Meadows' music policy?
Part of our main focus from early on was to promote underground music rather than have big headliners, to be taste-makers and find breakthrough acts who haven't been showcased. From early on we knew that the festival market was saturated, with lots of festivals essentially doing the same thing, and we want to be the opposite of that, so that needed to come across in the bookings. Plus, budget-wise we could never really afford to book big acts, so we went the other way.
What're your plans for the summer?
We're hoping to be involved in the Eclipse Festival in Oregon, which is by the same guys who run the Symbiosis Gathering and Burning Man. They follow the solar eclipse and invite all of the best festivals in the world to collaborate and create this epic festival.
Do you and Damian now work solely on festivals?
I still work as a civil engineer and have worked on festivals like Wilderness, Burning Man, Secret Garden Party. I just got back from Soysambu in the Wild, which generates money to try to save all the lions, giraffes and local wildlife. I also run a live music venue called Folklore on Hackney Road. Damian's got his own events planning and set production company called The Bearded Butler and does a lot of builds at other festivals, London Fashion Week and big exhibitions too.
You travel a lot – what would be your ideal way to spend a free weekend back home in London?
I don't really have a social life anymore, so seeing friends would be a priority! I don't go to big clubs much now but prefer getting around a table with my best friends, some food and good wine. London is so culturally rich and I enjoy going to live gigs here, but I'd also love to get out of the city and rent a cottage in the middle of nowhere for the weekend – hopefully with a friend who has a dog.
It was a book about the Bulgarian people, and how strong the empire used to be. It's so rich, all the history, and I'd love to spend more time researching that. We've never really had a theme for Meadows but I'm hoping that we might be able to do something with this idea of Bulgarian folklore – it's quite Pagan and there are references to the idea that the Meadows site used to be considered the land of the dragons.


The orientation program will provide you with first-hand experience in understanding how the educational system operates in China. It also gives you an opportunity to observe some typical classroom situations. Graduates of Appalachians Abroad Teach in China Program will share their experience on living and teaching in China. They will also give you concrete suggestions for responding most effectively to the many aspects of Chinese life which you will encounter inside and outside of your classrooms.
Some local sight seeing and cultural activities will be arranged. Accommodations and local transportation during the orientation program in Shanghai are included in the program fee.


text:
\section{Introduction}
\label{sec:1}

\vspace*{1mm}
\noindent
At the planned future $e^+e^-$ facilities which operate at high energy and at large luminosity, like
the ILC, CLIC \cite{ILC,Aihara:2019gcq,Mnich:2019,CLIC}, the FCC\_ee \cite{FCCEE}, and also muon colliders
\cite{Delahaye:2019omf}, tests of the Standard Model are possible at unprecedented accuracy. This concerns
a further detailed exploration of the $Z$ peak, beyond what was possible at LEP \cite{ALEPH:2005ab},
detailed studies of Higgs boson production using $Z H$ final states, accurate scans of the top-quark
threshold region, and various other precision measurements. Due to this the accuracy of the masses and
widths of the heaviest particles of the Standard Model will be significantly improved.

One important ingredient to these experimental precision studies are the QED radiative corrections and in
particular those due to initial state radiation (ISR). Very recently the $O(\alpha^2)$ ISR corrections for the
process $e^+e^- \rightarrow \gamma^*/Z^*$ have been completed in a direct calculation~\cite{Blumlein:2019srk,
Blumlein:2019pqb,QED2019}. Here $\alpha$ denotes the fine structure constant. It has been shown that the
result for all channels of a previous calculation
\cite{Berends:1987ab} needed to be corrected in the non--logarithmic terms at $O(\alpha^2)$. Agreement has
been found with the results of Ref.~\cite{Blumlein:2011mi}. Due to this it has also been proven that one may
use the method of massive operator matrix elements (OMEs) for these calculations and that the Drell--Yan process
factorizes for massive fermionic states.

In the present paper we take advantage of this method and extend the calculation to the first three
logarithmic terms up to the order $O(\alpha^5)$  beyond the complete $O(\alpha^2)$ corrections and thus reach
$O(\alpha^5 L^3)$. For comparison we also calculate the leading order contributions of $O(\alpha^6 L^6)$ and one
more subleading term $O(\alpha^6 L^5)$, with
$L = \ln(s/m_e^2)$, $s$ denotes the center of mass energy
squared of the annihilation process and $m_e$ is the electron mass. The universal corrections $O((\alpha L)^k)$
are known in analytic form to order  $k = 5$, cf.~\cite{Skrzypek:1992vk,Jezabek:1992bx,Przybycien:1992qe,
Blumlein:1996yz,Arbuzov:1999cq,Arbuzov:1999uq,Blumlein:2004bs,Blumlein:2007kx}, accounting for the non--singlet
and singlet contributions in the unpolarized and polarized case. The method used in
Ref.~\cite{Blumlein:2011mi} can be extended to higher orders in the coupling constant. For the first
subleading term at $O(\alpha^3)$\footnote{Subleading corrections to differential cross sections have also been
studied, cf.~\cite{Blumlein:2002fy}.}, all ingredients forming the
radiators in terms of Mellin transforms are known from the calculation of the anomalous dimensions in QCD
\cite{NLOand}, the Wilson coefficients of the massless Drell--Yan process
\cite{Hamberg:1990np,Harlander:2002wh,Blumlein:2019srk} and the massive OMEs in \cite{Blumlein:2011mi}. To
obtain the $O(\alpha^3 L)$ and $O(\alpha^3 L)$ correction we need also to calculate the massive OME
$\Gamma^{(1)}_{\gamma e}$.
This also applies to the higher order subleading contributions. This series could be continued straightforwardly to
higher and higher order, for the first three terms at each order in the logarithmic expansion in Mellin $N$ space
on the expense of longer and longer expressions.

It turns out that it is indeed the case that one has to reach at least corrections of the order $O(\alpha^5 L^4)$
for the ISR
corrections to satisfy the ambitions goals at the FCC\_ee of $\sim 100$~keV both for the $Z$ mass, $M_Z$, and the $Z$
width, $\Gamma_Z$, on the theoretical side. The method of massive operator matrix elements \cite{Blumlein:2011mi} makes
this calculation possible. In the present approach the constant term of $O(\alpha^3)$ and its higher order logarithmic
extensions are still missing. They require still higher order massive OMEs and also massless Wilson
coefficients in analytic form. Their size is, however, gradually smaller.

The paper is organized as follows. In Section~\ref{sec:2} we present the structure of the QED ISR radiative
corrections to the Born cross section following from the renormalization group equations (RGEs), for which we present
the analytic solutions in terms of anomalous dimensions, massive OMEs and massless Wilson coefficients for all
contributions calculated in the present paper. In Section~\ref{sec:3} we calculate the massive OME
$\Gamma^{(1)}_{\gamma e}$. The radiator functions in $z$ space for the $O(\alpha^3)$ and $O(\alpha^4)$
corrections are presented
in Section~\ref{sec:4}.  The radiators can be expressed in terms of harmonic polylogarithms \cite{Remiddi:1999ew},
to which the Nielsen integrals and classical polylogarithms form a subset \cite{NIELSEN,CLPOLY}.
In Section~\ref{sec:5}, we present numerical results on the corrections in the kinematic
region around the $Z$ peak and we determine the corresponding corrections to the pole mass of the
$Z$ boson and the $Z$
boson width, $\Gamma_Z$. Section~\ref{sec:6} contains the conclusions. In Mellin $N$ space the radiators are given in
terms of harmonic and generalized harmonic sums. If compared to the $z$ space representation they are more compact.
Due to the appearance of generalized harmonic sums we derive in Appendix~\ref{sec:A} the singularity structure of
the radiators in the complex $N$-plane and in Appendix~\ref{sec:B} we present all radiators
calculated in the present paper in Mellin space.
\section{The Initial State Corrections to the \boldmath $e^+e^-$ Annihilation Cross Section}
\label{sec:2}

\vspace*{1mm}
\noindent
The initial state QED radiative corrections to $e^+e^- \rightarrow \gamma^*/Z^*$ can be calculated
by applying the method of massive operator matrix elements, cf.~\cite{Blumlein:2011mi}. It  has been demonstrated
by the recent complete calculation in \cite{Blumlein:2019srk,Blumlein:2019pqb,QED2019} that the effective method
of Ref.~\cite{Blumlein:2011mi} is delivering the complete result. Particular non--logarithmic contributions with
vanishing massive OME to the Drell--Yan process in the constant term $O(\alpha^2)$ could be structurally absorbed
in the relations and appear as contributions to the massless Wilson coefficients given in
\cite{Blumlein:2011mi},~Eqs.~(40--42). It is due to this agreement, that one
can now safely apply this method also for subleading logarithms to higher orders in the coupling constant by
solving the associated renormalization group equations in the massive case.
The method has been used in Ref.~\cite{Berends:1987ab} before for the logarithmic enhanced contributions
to $O(\alpha^2 L)$.\footnote{Note
the necessary corrections of relations in \cite{Berends:1987ab} given in Ref.~\cite{Blumlein:2011mi}.}

The massive effects due to the finite electron mass $m_e$ enter here through process--independent massive operator
matrix
elements. The Wilson coefficients are those of the massless Drell--Yan process \cite{Hamberg:1990np,Harlander:2002wh,
Blumlein:2019srk}. Note that there are differences between the corrections to the vector and axial--vector coupling.

In $z$--space the different contributions to the radiators are connected by Mellin convolutions $\otimes$ which
are defined by
\begin{eqnarray}
[A \otimes B](z) = \int_0^1 dx_1 \int_0^1 dx_2 \delta(1 - x_1 x_2) A(x_1) B(x_2).
\end{eqnarray}
The Mellin transform reads
\begin{eqnarray}
\label{eq:MELTRA}
{\rm\bf M}[f(z)](N) = \int_0^1 dz z^{N-1} f(z),~~~~
{\rm\bf M}[[f(z)]_+](N) = \int_0^1 dz (z^{N-1}-1) f(z)
\end{eqnarray}
for regular functions and +-distributions, respectively.
The most recently calculated quantities are the massive OMEs given in \cite{Blumlein:2011mi}.

The radiative corrections to the differential scattering cross section is given by
\begin{eqnarray}
\label{EQ:SI1}
\frac{d\sigma_{e^+e^-}}{ds'} = \frac{1}{s} \sigma^{(0)}\left\{1 + \sum_{k = 1}^\infty a_0^k \sum_{l=0}^k c_{k, l}
{\rm \bf L}^{k-l} \right\}
\end{eqnarray}
in $z$ space. Here we defined
\begin{eqnarray}
{\rm \bf L} =
L  + \ln(z),~~~~~L = \ln \left(\frac{s}{m_e^2}\right),
\end{eqnarray}
and $\sigma^{(0)}$ denotes the Born cross section, cf.~\cite{Blumlein:2011mi},~Eq.~(8), and Ref.~\cite{BDJ}.
$a_0 = a(\mu^2 = m_e^2)$ is the normalized fine structure constant with $a(\mu^2) = \alpha(\mu^2)/(4\pi)$, which
we will widely use in the following. Here it is convenient to refer to $a_0$ only and account also for all
evolution contributions by the functions $c_{k,l}$.

For the Born cross section for $e^+e^-$ annihilation, $\sigma_0$, we  will consider
$s$-channel $e^+e^-$ annihilation into a virtual gauge boson $(\gamma, Z)$ which decays into a fermion pair
$f \overline{f}$ and $e \neq f$. This process both describes lower energy $\gamma$-exchange and the $Z$-resonance.
\begin{eqnarray}
\label{eq:BO1}
\frac{d\sigma^{(0)}(s)}{d \Omega} &=& \frac{\alpha^2}{4 s}
                                        N_{C,f} \sqrt{1 - \frac{4 m_f}{s}}
\nonumber\\
& & \times
\left[\left(1+ \cos^2\theta + \frac{4 m_f^2}{s} \sin^2\theta \right) G_1(s)
- \frac{8 m_f^2}{s} G_2(s)
+ 2 \sqrt{1-\frac{4m_f^2}{s}} \cos\theta G_3(s)\right]
\nonumber\\
&& \times {\cal G}(s)~,
\\
\label{eq:BO2}
\sigma^{(0)}(s) &=& \frac{4 \pi \alpha^2}{3 s}
                                        N_{C,f} \sqrt{1 - \frac{4 m_f}{s}}
\left[\left(1 + \frac{2 m_f^2}{s} \right) G_1(s)
- 6 \frac{m_f^2}{s} G_2(s)\right] {\cal G}(s)~,
\end{eqnarray}
see e.g. \cite{BDJ,WB}\footnote{Note a missing term in \cite{Berends:1987ab}, Eq.~(2.5).}.
Here the final state fermions are considered to be no electrons, to obtain an $s$-channel Born cross section.
In
Eqs.~(\ref{eq:BO1}, \ref{eq:BO2}) the electron mass is neglected kinematically. $N_{C,f}$ is the number of colors
of the final state fermion, with $N_{C,f} = 1$ for colorless fermions,
and $N_{C,f} = 3$ for quarks. The function ${\cal G}(s) = 1$ in the case of the pure perturbative calculation.
$s$ is the cms energy, $\Omega$ is the spherical angle, $\theta$ the cms scattering
angle, and the effective couplings $G_i(s)|_{i=1...3}$ read
\begin{eqnarray}
G_1(s) &=& Q_e^2 Q_f^2 + 2 Q_e Q_f v_e v_f {\sf Re}[\chi_Z(s)]
          +(v_e^2+a_e^2)(v_f^2+a_f^2)|\chi_Z(s)|^2,\\
G_2(s) &=& (v_e^2+a_e^2) a_f^2 |\chi_Z(s)|^2, \\
G_3(s) &=& 2 Q_e Q_f a_e a_f {\sf Re}[\chi_Z(s)] + 4 v_e v_f a_e a_f |\chi_Z(s)|^2.
\end{eqnarray}
The reduced $Z$--propagator is given by
\begin{eqnarray}
\chi_Z(s) = \frac{s}{s-M_Z^2 + i M_Z \Gamma_Z},
\end{eqnarray}
where $M_Z$ and  $\Gamma_Z$ are the mass and the width of the $Z$ boson and $m_f$ is the mass of the final
state fermion. $Q_{e,f}$ are the electromagnetic charges of the electron $(Q_e = -1)$
and the final state fermion, respectively, and
the electro--weak couplings $v_i$ and $a_i$ read
\begin{eqnarray}
v_e &=& \frac{1}{\sin\theta_w \cos\theta_w}\left[I^3_{w,e} - 2 Q_e
\sin^2\theta_w\right],\\
a_e &=& \frac{1}{\sin\theta_w  \cos\theta_w} I^3_{w,e}, \\
v_f &=& \frac{1}{\sin\theta_w \cos\theta_w}\left[I^3_{w,f} - 2 Q_f
\sin^2\theta_w\right],\\
a_f &=& \frac{1}{\sin\theta_w  \cos\theta_w} I^3_{w,f}~,
\end{eqnarray}
where $\theta_w$ is the weak mixing angle, and $I^3_{w,i} = \pm 1/2$ the third component
of the weak isospin for up and down particles, respectively.

The inclusive $s$--channel annihilation scattering cross section $\sigma_{e^+ e^-}$ is given by
\begin{eqnarray}
\label{eq:incl}
\sigma_{e^+ e^-}(s) &=& \int_{s_0}^s ds' \frac{d \sigma_{e^+ e^-}(s')}{ds'}
= \int_{z_0}^1 dz \sigma^{(0)}(s z) R(z,s/m_e^2),
\end{eqnarray}
with $s' = sz$, $\sigma^{(0)}(s')$ the Born cross section and $R(z,s/m_e^2)$ the distribution--valued radiator
\cite{DISTR} describing the initial state radiation of photons and light $e^+e^-$ pairs. The lower bound $s_0 =
s z_0$ is an invariant mass squared depending on the experiment, cf.~\cite{ALEPH:2005ab}. In the later numerical
illustrations we will choose $s_0 = 4 m_\tau^2$, with $m_\tau$ the $\tau$ lepton mass.

The general decomposition of the scattering cross section in Mellin space is given by, cf.~\cite{Berends:1987ab}\footnote{
In the massless case the principle solution of the RGEs to general
orders has been known for long, see~\cite{Blumlein:1997em,Ellis:1993rb}.}
\begin{eqnarray}
\label{EQ:SI2}
\frac{d\sigma_{e^+e^-}}{ds'} &=& \frac{1}{s} \sigma^{(0)}(s')
\Biggl[ \Gamma_{e^+e^+}\left(N,\frac{\mu^2}{m_e^2}\right)
\tilde{\sigma}_{e^+e^-}\left(N,\frac{s'}{\mu^2}\right)
\Gamma_{e^-e^-}\left(N,\frac{\mu^2}{m_e^2}\right)
\nonumber\\
&& +
\Gamma_{\gamma e^+}\left(N,\frac{\mu^2}{m_e^2}\right)
\tilde{\sigma}_{e^- \gamma}\left(N,\frac{s'}{\mu^2}\right)
\Gamma_{e^-e^-}\left(N,\frac{\mu^2}{m_e^2}\right)
\nonumber\\
&& +
\Gamma_{e^+ e^+}\left(N,\frac{\mu^2}{m_e^2}\right)
\tilde{\sigma}_{e^+ \gamma}\left(N,\frac{s'}{\mu^2}\right)
\Gamma_{\gamma e^-}\left(N,\frac{\mu^2}{m_e^2}\right)
\nonumber\\
&& +
\Gamma_{\gamma e^+}\left(N,\frac{\mu^2}{m_e^2}\right)
\tilde{\sigma}_{\gamma \gamma}\left(N,\frac{s'}{\mu^2}\right)
\Gamma_{\gamma e^-}\left(N,\frac{\mu^2}{m_e^2}\right)
\Biggr].
\end{eqnarray}
The terms in the brackets $[...]$ are Mellin--convoluted. Only massive OMEs of the kind
$\Gamma_{e^{\pm} e^{\pm}}$ and $\Gamma_{\gamma e^{\pm}}$
contribute because the process considered has electron--positron initial states.
The last term in Eq.~(\ref{EQ:SI2}) is only contributing with $O(a^4)$. $\mu$ denotes the
factorization and renormalization scale. As we will see later, it will cancel in the scattering cross section,
when performing the expansion consistently to a certain order in $a_0$.
The massive OMEs, $\Gamma_{ij}$, and massless Wilson coefficients, $\tilde{\sigma}_{ij}$, obey the following
series representations
\begin{eqnarray}
\Gamma_{li}\left(N, \frac{\mu^2}{m_e^2}\right)       &=& \delta_{li} + \sum_{r=1}^\infty a^r(\mu^2) \sum_{n=0}^r
a_{li;nr}(N) \Lambda^n
\\
\tilde{\sigma}_{lk}\left(N, \frac{s'}{\mu^2} \right) &=& \delta_{lk} + \sum_{r=1}^\infty a^r(\mu^2) \sum_{n=0}^r
b_{lk;nr}(N)
\lambda^n,
\end{eqnarray}
with the logarithms $\Lambda$ and $\lambda$ given by
\begin{eqnarray}
\Lambda        = \ln\left(\frac{\mu^2}{m_e^2}\right),~~~~~~~
\lambda  = \ln\left(\frac{s'}{\mu^2}\right).
\end{eqnarray}
The massive OMEs $\Gamma_{ij}$ and massless Wilson coefficients $\tilde{\sigma}_{kl}$ fulfill the
following renormalization group equations, cf.~\cite{Blumlein:2000wh},
\begin{eqnarray}
\label{eq:RE1}
\left[\left(\frac{\partial}{\partial \Lambda} + \beta(a) \frac{\partial}{\partial a}\right) \delta_{kl} +
\frac{1}{2}
\gamma_{kl}(N,a)\right]
\Gamma_{li}\left(N,a,\frac{\mu^2}{m_e^2}\right)
&=& 0 \\
\label{eq:RE2}
\left[\left(\frac{\partial}{\partial \lambda} - \beta(a) \frac{\partial}{\partial a}\right) \delta_{kl} \delta_{jm}
+ \frac{1}{2} \gamma_{kl}(N,a) \delta_{jm}
+ \frac{1}{2} \gamma_{jm}(N,a) \delta_{kl} \right]
\tilde{\sigma}_{lj}\left(N,a,\frac{s'}{\mu^2}\right) &=& 0~,
\end{eqnarray}
and the QED $\beta$ function has the representation
\begin{eqnarray}
\beta(a) = - \sum_{k=0}^\infty \beta_k a^{k+2}.
\end{eqnarray}
Eq.~(\ref{eq:12}) gives an overview on the orders of the expansion of the radiators in the fine
structure constant which are now available, including the results of the present calculation,
\begin{equation}
\label{eq:12}
\begin{array}{lll}
\alpha \ {L}   &  \alpha                 &
\\
\alpha^2 { L}^2 &  \alpha^2 {L}   & \alpha^2
\\
\alpha^3 {L}^3 &  \alpha^3 {L}^2 & \alpha^3 {L}
\\
\alpha^4 {L}^4 &  \alpha^4 {L}^3 & \alpha^4 {L}^2
\\
\alpha^5 {L}^5 &  \alpha^5 {L}^4 & \alpha^5 {L}^3
\\
\alpha^6 {L}^6 &  \alpha^6 {L}^5~.   &
\end{array}
\end{equation}
The expansion coefficients $c_{k, l}$ of Eq.~(\ref{EQ:SI1}) up to the sixth order in $a_0$ in Mellin space are
given by Eqs.~(\ref{EQ:EXP1}--\ref{EQ:EXP21}). They are expressed by the anomalous dimensions
$\gamma_{ij}^{(k)}$
\cite{NLOand}, the expansion
coefficients of the massless Drell--Yan cross section \cite{Hamberg:1990np,Harlander:2002wh,Blumlein:2019srk},
$\tilde{\sigma}_{ij}^{(k)}$, the expansion coefficients of the QED $\beta$ function and the renormalized massive OMEs
$\Gamma_{ij}^{(k)}$, where $(k+1)$ denotes the
loop order. In the following we use the notation
\begin{eqnarray}
 \gamma_{ij}^{(k)} = - P_{ij}^{(k)}(N) = - {\rm\bf M}[P_{ij}^{(k)}(z)](N).
\end{eqnarray}
When working in $z$ space we will use $P_{ij}^{(k)}(z)$ instead of the anomalous dimensions
$\gamma_{ij}^{(k)}$.

One obtains
\input{formu2.tex}
where $P_{\gamma\gamma}^{(0)} = - 2 \beta_0$.
These functions are derived by solving the renormalization group equations for the massive OMEs and
the massless Wilson coefficients, (\ref{eq:RE1}, \ref{eq:RE2}) up to six--loop order. We show also the
constant term $c_{3,0}$ which contains the term $\Gamma_{ee}^{(2)}$, not having been calculated yet.\footnote{
For QCD the Drell--Yan cross section $\tilde{\sigma}^{(2)}$ has been  calculated
giving numerical illustrations in \cite{Duhr:2020seh} very recently.}

A few remarks are in order. The sub--system cross section for the Drell--Yan process is flavor dependent,
cf.~Eq.~(\ref{EQ:SI2}), which results in the present case from the to $e^+ e^-$ initial state, the vertex to which the
produced neutral current gauge bosons, $\gamma^*$
or $Z^*$, couple. I.e. e.g. in the term describing intermediate photon exchange in
Eq.~(\ref{eq:FF2}) by $P_{e\gamma}^{(0)} P_{\gamma e}^{(0)}$ has to be read as either
$P_{e^-\gamma}^{(0)} P_{\gamma e^-}^{(0)}$ or
$P_{e^+\gamma}^{(0)} P_{\gamma e^+}^{(0)}$. Therefore, one has
\begin{eqnarray}
P_{e^- \gamma}^{(0)}(z) = P_{e^+ \gamma}^{(0)}(z) = 4 [z^2 + (1-z)^2],
\end{eqnarray}
and the energy--momentum sum--rule
\begin{eqnarray}
\label{eq:sr1}
\int_0^1 dz~z~[P_{ee}^{(0)}(z) + P_{e^- \gamma}^{(0)}(z) + P_{e^+ \gamma}^{(0)}(z)] = 0,
\end{eqnarray}
is obeyed. Similar relations hold in higher order and also apply to the corresponding massive OMEs.
$P_{ee}^{(0)}$ is flavor conserving, i.e. it does either describe an $e^- \rightarrow e^-$ or an $e^+ \rightarrow e^+$
transition.

Furthermore, one has
\begin{eqnarray}
\label{eq:sr2}
\int_0^1 dz~z~[P_{\gamma e}^{(0)}(z) + P_{\gamma \gamma}^{(0)}(z)] = 0,
\end{eqnarray}
where $P_{\gamma e}^{(0)}(z) = P_{\gamma e^+}^{(0)}(z) + P_{\gamma e^-}^{(0)}(z)$, like in Quantum Chromodynamics
(QCD) setting the color factor $C_F = 1$, cf.~\cite{Blumlein:2012bf}. This complication is usually not present in QCD,
since there the splitting functions act in the singlet case on fully symmetrized flavor distributions, like the flavor
singlet distribution $\Sigma(x,Q^2)$, summing over all quark and antiquark flavors.

In the above, $P_{ij}^{(k)}$ denotes the Mellin transform of the  corresponding expansion coefficient  of the
splitting function. The building blocks for the above quantities are given in Eqs.~(80--82, 90--92, 94, 95) of
Ref.~\cite{Blumlein:2011mi} and Refs.~\cite{Hamberg:1990np,Harlander:2002wh,Blumlein:2019srk,NLOand}
in $z$ space and the operator matrix element calculated in Section~\ref{sec:3}.   Furthermore, a series of
Mellin convolutions is needed which are given in Appendix~A of
\cite{Blumlein:2011mi}. Further higher order convolutions can be carried out by the algorithms encoded in
the package {\tt HarmonicSums}~\cite{Vermaseren:1998uu,Blumlein:1998if,Ablinger:2014rba,Ablinger:2010kw, Ablinger:2013hcp,
Ablinger:2011te,
Ablinger:2013cf,Ablinger:2014bra,Ablinger:2017Mellin}.

The running coupling constant is the solution of the differential equation
\begin{eqnarray}
\label{EQ:run1}
\frac{d a(\mu^2)}{d\ln(\mu^2)} = - \sum_{k=0}^\infty \beta_k a^{k+2}(\mu^2),
\end{eqnarray}
with $\beta_k$ the expansion coefficients of the $\beta$-function, with
\begin{eqnarray}
\label{EQ:runb}
\beta_0 = -\frac{4}{3},~~~~~\beta_1 = -4,~~~~~\beta_2 = \frac{62}{9},
\end{eqnarray}
\cite{Gross:1973id,Politzer:1973fx,Caswell:1974gg,Vladimirov:1979zm} in the single fermion approach, $N_F = 1$,
retaining only electrons. Since we are dealing with the first three logarithmic orders from $O(\alpha^3)$ onward
only terms up to $\beta_2$ are contributing. The solution of (\ref{EQ:run1}) has been derived in
Ref.~\cite{Blumlein:1994kw}, Eqs.~(2, 3),  in the $\overline{\sf MS}$ scheme by keeping all terms up to $\beta_2$
in closed form. The corresponding perturbative solution for $a(\mu^2) = a(a_0,L)$ is then given by
\begin{eqnarray}
\label{EQ:run3}
a(\mu^2) = a_0 - a_0^2 L \beta_0
         + a_0^3 L \left(\beta_0^2 L   - \beta_1\right)
         - a_0^4 L \left(\beta_0^3 L^2 - \frac{5}{2} \beta_0 \beta_1 L + \beta_2\right) + O\left(a_0^5\right).
\end{eqnarray}
One verifies the correctness of (\ref{EQ:run3}) by inserting it into Eq.~(\ref{EQ:run1}).
Note that if one does not expand the fine structure constant $a(\mu^2)$ w.r.t. its reference value $a_0 = a(\mu^2 =
m_e^2)$, the $\mu$ dependence in the RGE--decomposition given in Ref.~\cite{Blumlein:2011mi} is not canceled.
However, expanding to the respective order in the coupling constant intended, a scheme--invariant expression is
obtained. This is the reason, why we are expanding the logarithmic dependence of the coupling constant and
write the scattering cross section in terms of $a_0$.

In the above expressions we presented the sub--system cross sections $\tilde{\sigma}_{ij}$ in a genuine way. One should
note that these quantities are partly different for vector and axial--vector couplings, cf.~\cite{Hamberg:1990np,QED2019}
and so are some of the radiators. The first expressions for $c_{i,j}$ in $z$ space are presented in
Section~\ref{sec:4}. In Appendix~\ref{sec:B} all coefficients $c_{i,j}$ in Mellin $N$ space, which are more compact,
are given.

Sumrules do not only hold for the splitting functions (\ref{eq:sr1}--\ref{eq:sr2}) but also for the universal
unrenormalized massive operator matrix elements, cf.~\cite{Buza:1996wv,BBK2} to two--loop order
\begin{eqnarray}
\int_0^1 dx  x\Biggl[
\hat{A}_{ee}^{\rm NS}  + \hat{A}_{ee}^{\rm PS} + \hat{A}_{\gamma e} \Biggr] &=& 1,
\end{eqnarray}
which we have verified for $N_F = 1$.
\section{The Operator Matrix Element \boldmath $\Gamma^{(1)}_{\gamma e}$}
\label{sec:3}

\vspace*{1mm}
\noindent
For the calculation of the operator matrix element $\Gamma^{(1)}_{\gamma e}$ we follow the notation of~\cite{Blumlein:2011mi}.
After wave function and mass renormalization we can write the operator matrix element as
\begin{eqnarray}
      \hat{A}_{\gamma e} &=& \hat{a} \cdot \hat{\hat{A}}_{\gamma e}^{(1)}
      + \hat{a}^2 \left[ \hat{\hat{A}}_{\gamma e}^{(2)} + Z_\text{CT}^{(2)} \right] + O(\hat{a}^3),
\end{eqnarray}
where $Z_\text{CT}^{(2)}$ are the counter term contributions due to mass and wave function renormalization,
$\hat{\hat{A}}_{\gamma e}^{(i)}$ are the unrenormalized operator matrix elements at $i$-loops and $\hat{a}$
is the unrenormalized fine structure constant.

The renormalized operator matrix elements in the $\text{MOM}$-scheme are given by, cf.~\cite{BBK2},
\begin{eqnarray}
      A_{\gamma e}^\text{MOM} &=& a^\text{MOM} \left[ \hat{A}_{\gamma e}^{(1)} + Z_{\gamma e}^{-1,(1)} \right]
      + \left. a^\text{MOM} \right.^2 \Bigl[ \hat{A}_{\gamma e}^{(2)} + Z_{\gamma e}^{-1,(2)} +  Z_{\gamma e}^{-1,(1)} \hat{A}_{e e}
\nonumber \\ &&
      + Z_{\gamma \gamma}^{-1,(1)} \hat{A}_{\gamma e} + \delta a_1^\text{MOM} \hat{A}_{\gamma e}^{(1)} \Bigr] + O( \left. a^\text{MOM} \right.^3) ,
\end{eqnarray}
where the unrenormalized coupling constant can be expressed via
\begin{eqnarray}
      \hat{a} &=& a^\text{MOM} \left[ 1 + \delta a_1^\text{MOM} a^\text{MOM} \right] + O\left(\left. a^\text{MOM}
\right.^3\right)
\end{eqnarray}
with
\begin{eqnarray}
      \delta a_1^\text{MOM} = S_\epsilon \frac{2 \beta_0}{\epsilon} \left( \frac{m^2}{\mu^2} \right)^{\epsilon/2}
      \exp\left[ \sum_{i=0}^\infty \frac{\zeta_i}{i} \left( \frac{\epsilon}{2} \right)^i \right],
\end{eqnarray}
and the spherical factor $S_\epsilon = \exp[(\epsilon/2)[\ln(4\pi) - \gamma_E]]$.

The operator matrix element, after charge and wave function renormalization, can therefore be written as
\begin{eqnarray}
      \hat{A}_{\gamma e} &=& a^\text{MOM} S_\epsilon \left( \frac{m^2}{\mu^2} \right)^{\epsilon/2} \left[ -\frac{1}{\epsilon} P_{\gamma e}^{(0)} + \Gamma_{\gamma e}^{(0)} + \epsilon \bar{\Gamma}_{\gamma e}^{(0)} \right]
      + \left. a^\text{MOM} \right.^2 S_\epsilon^2 \left( \frac{m^2}{\mu^2} \right)^{\epsilon} \Biggl\{
         \frac{\gamma_{\gamma e}^{(0)}}{2\epsilon^2}
         \bigl[
             \gamma_{ee}^{(0)} + \gamma_{\gamma\gamma}^{(0)} - 4 \beta_0
         \bigr]
\nonumber \\ &&
       + \frac{1}{\epsilon}
         \biggl[
            \frac{1}{2} \gamma_{\gamma e}^{(1)}
            + \gamma_{\gamma e}^{(0)} \Gamma_{ee}^{(1)}
            + \gamma_{\gamma \gamma}^{(0)} \Gamma_{\gamma e}^{(1)}
            - 2 \beta_0 \Gamma_{\gamma e}^{(1)}
         \biggr]
       + \hat{\Gamma}_{\gamma e}^{(1)} \Biggr\}~.
\end{eqnarray}
The predicted pole structure serves as a test on the calculation.
The renormalized OME in the $\overline{\sf MS}$-scheme is given by
\begin{eqnarray}
      A_{\gamma e} &=& a^{\overline{\sf MS}} \left[ - \frac{1}{2} P_{\gamma e}^{(0)} L + \Gamma_{\gamma e}^{(0)} \right]
      + \left. a^{\overline{\sf MS}} \right.^2 \Bigl[
      \frac{P_{\gamma e}^{(0)}}{8} \left( P_{ee}^{(0)} + P_{\gamma\gamma}^{(0)} + 2 \beta_0 \right) L^2
\nonumber \\ &&
      + \frac{1}{2} \left( P_{\gamma e}^{(1)}
      + \Gamma_{e e}^{(0)} P_{\gamma e}^{(0)}
      + \Gamma_{\gamma e}^{(0)} P_{\gamma\gamma}^{(0)}
      + 2 \beta_0 \Gamma_{\gamma e}^{(0)}
      \right) L
      + \hat{\Gamma}_{\gamma e}^{(1)}
\nonumber \\ &&
      + \bar{\Gamma}_{ee}^{(0)} P_{\gamma e}^{(0)}
      + \bar{\Gamma}_{\gamma e}^{(0)} P_{\gamma \gamma}^{(0)}
      + 2 \beta_0 \bar{\Gamma}_{\gamma e}^{(0)}
      \Bigr],
\end{eqnarray}
where we used the relation
\begin{eqnarray}
      a^{\text{MOM}} = a^{\overline{\sf MS}} + \beta_0 L \left. a^{\overline{\sf MS}} \right.^2
\end{eqnarray}
and
\begin{eqnarray}
\hat{\Gamma}_{ij} = {\Gamma}_{ij}(N_F+1) - {\Gamma}_{ij}(N_F).
\end{eqnarray}

The Feynman diagrams contributing to $\Gamma^{(1)}_{\gamma e}$ are shown in Figure~\ref{DIAG1}, with the
corresponding
symmetrization understood. They can be represented in terms of 18 master integrals by performing the
integration-by-parts
reduction using the package {\tt Litered} \cite{LR}. Here, as in
previous calculations,
cf.~\cite{Ablinger:2014vwa}, we resummed the local operators into linear
propagators. The master integrals are either calculated using conventional methods like generalized hypergeometric
functions, cf. e.g. \cite{Blumlein:2018cms}, or can be obtained by solving ordinary differential equations, cf.
e.g.~\cite{Ablinger:2018zwz}.\footnote{We took the opportunity to re--calculate the results of \cite{Blumlein:2011mi}
by using the same techniques. Here also 18 master integrals contribute, which can be calculated in a similar manner.
This can now be done in a fully automated way. The large mathematical and conceptional progress in performing loop
integrals since 2002 is clearly demonstrated by this.}
The code {\tt Tarcer} \cite{Mertig:1998vk} has been used for checks and to determine
initial values.
\begin{figure}[H]
\centering

\vspace*{-1cm}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f1-eps-converted-to.pdf}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f2-eps-converted-to.pdf}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f3-eps-converted-to.pdf}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f4-eps-converted-to.pdf}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f5-eps-converted-to.pdf}

\vspace*{-17cm}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f6-eps-converted-to.pdf}
\includegraphics[width=0.19\linewidth,height=20cm]{figs/f7-eps-converted-to.pdf}

\vspace*{-17cm}
\caption{\sf
The Feynman diagrams contributing to $\Gamma^{(1)}_{\gamma e}$. For the notation and the Feynman rules see
Ref.~\cite{Blumlein:2011mi}. The symbol {\tiny $\bigstar$} denotes the counter-term insertion.}
\label{DIAG1}
\end{figure}
\noindent

The expansion coefficients of the unrenormalized OME $\hat{A}_{\gamma e}^{(1)}$ are given by
\begin{eqnarray}
\hat{A}_{\gamma e}^{(1)} &=&
\hat{a} \left(\frac{m_e^2}{\mu^2}\right)^{\epsilon/2} S_\epsilon \left[
- \frac{1}{\epsilon} P_{\gamma e}^{(0)}(z)
+ \Gamma_{\gamma e}^{(0)}(z)
+ \epsilon \overline{\Gamma}_{\gamma e}^{(0)}(z) + O(\epsilon^2) \right],
\end{eqnarray}
see \cite{Blumlein:2011mi}.
The coefficient $\Gamma_{\gamma e}^{(0)}$ reads
\begin{eqnarray}
\Gamma_{e \gamma}^{(0)} &=& -2 \frac{1 + (1-z)^2}{z}\left[2 \ln(z) +1\right].
\end{eqnarray}

The corresponding result in Mellin $N$ space is the obtained as the $N$th expansion coefficient in the auxiliary variable
$\hat{x}$ and the $z$ space representation can be obtained by a subsequent inverse Mellin transform.
The renormalized OME $A_{\gamma e}^{(2)}$ is given by
\begin{eqnarray}
\label{eq:AN}
      A_{\gamma e}^{(2)}(N) &=&
      \Biggl[
            \frac{\big(N^2+N+2\big)\big(N^2+N+6\big)}{3 (N-1) N^2 (N+1)^2}
            -\frac{4 \big(N^2+N+2\big)}{(N-1) N (N+1)} S_1
      \Biggr] L^2
\nonumber \\ &&
      - \Biggl[
            \frac{2 P_2}{9 (N-1)^2 N^3 (N+1)^3}
            -\frac{4 P_1}{3 (N-1) N^2 (N+1)^2} S_1
            +\frac{12 \big(N^2+N+2\big)}{(N-1) N (N+1)} S_1^2
\nonumber \\ &&
            +\frac{12 \big(N^2+N+2\big)}{(N-1) N (N+1)} S_2
      \Biggr] L
      +\frac{P_ 8}{27 (N-4) (N-3) (N-2) (N-1) N^4 (N+1)^4}
\nonumber \\ &&
      +\biggl(
            \frac{2 P_ 7}{9 (N-4) (N-3) (N-2) (N-1) N^3 (N+1)^3}
            +\frac{2 \big(N^2+N+2\big)}{(N-1) N (N+1)} S_ 2
      \biggr) S_ 1
\nonumber \\ &&
      +\frac{P_ 3}{3 (N-2) (N-1) N (N+1)^2} S_ 1^2
      +\frac{2 \big(N^2+N+2\big)}{3 (N-1) N (N+1)} S_ 1^3
\nonumber \\ &&
      +\frac{P_ 6}{3 (N-2) (N-1) N^2 (N+1)^2} S_ 2
      +\frac{4 \big(N^2+N+2\big)}{3 (N-1) N (N+1)} S_ 3
\nonumber \\ &&
      -\frac{48 \big(N^2+N+2\big)}{(N-1) N (N+1)} S_{2,1}
      +\frac{3 \cdot 2^{6+N}}{(N-2) (N+1)^2} S_ {1,1}\biggl(\frac{1}{2},1\biggr)
\nonumber \\ &&
      + \frac{2^{6-N} P_ 5}{3 (N-3) (N-2) (N-1)^2 N^2}
      \biggl(
              S_ 2(2)
            + S_ 1 S_1(2)
            - S_{1,1}(1,2)
            - S_{1,1}(2,1)
      \biggr)
\nonumber \\ &&
      -\frac{32 \big(N^2+N+2\big)}{(N-1) N (N+1)}
      \biggl[
            S_ 1(2) S_ {1,1}\biggl(\frac{1}{2},1\biggr)
            + S_ {1,2}\biggl(\frac{1}{2},2\biggr)
            - S_ {1,1,1}\biggl(\frac{1}{2},1,2\biggr)
\nonumber \\ &&
            - S_ {1,1,1}\biggl(\frac{1}{2},2,1\biggr)
            - \frac{\zeta_2}{2} S_1(2)
      \biggr]
      + \frac{4 P_ 4}{(N-2) (N-1) N^2 (N+1)^2} \zeta_2,
\end{eqnarray}
with the polynomials
\begin{eqnarray}
      P_1 &=& 25 N^4+44 N^3+87 N^2+56 N+12,
\\
      P_2 &=& 112 N^7+194 N^6+347 N^5+339 N^4+93 N^3-293 N^2-60 N+36,
\\
      P_3 &=& 17 N^4-66 N^3-179 N^2-272 N-212,
\\
      P_4 &=& N^5+4 N^4 +25 N^3+14 N^2+12 N+8 -3 \cdot 2^{N+3} N^2 \bigl(N - 1\bigr),
\\
      P_5 &=& 9 N^5-24 N^4+8 N^3+4 N^2+33 N-18,
\\
      P_6 &=& 11 N^5-90 N^4-329 N^3-356 N^2-284 N-48,
\\
      P_7 &=& 17 N^9+213 N^8-1729 N^7+2329 N^6-5196 N^5+7898 N^4+16196 N^3
\nonumber \\ &&
      +12528 N^2-4896 N-3456,
\\
      P_8 &=& -509 N^{11}+2365 N^{10}+2797 N^9-13158 N^8+31274 N^7-4694 N^6-64636 N^5
\nonumber \\ &&
      -107861 N^4-14622 N^3+6588 N^2-2376 N-2592.
\end{eqnarray}
$A_{\gamma e}^{(2)}$ is expressed by harmonic sums \cite{Vermaseren:1998uu,Blumlein:1998if}
\begin{eqnarray}
S_{b,\vec{a}}(N) = \sum_{k=1}^N \frac{({\rm sign}(b))^k}{k^{|b|}} S_{\vec{a}}(k),~~~S_\emptyset = 1, a_i,b_i \in
\mathbb{Z} \backslash \{0\},
\end{eqnarray}
and generalized harmonic sums \cite{Moch:2001zr,Ablinger:2013cf}
\begin{eqnarray}
S_{b,\vec{a}}(d,\vec{c},N) = \sum_{k=1}^N \frac{d^k}{k^{|b|}} S_{\vec{a}}(\vec{c},k),~~~S_\emptyset = 1,
a_i,b \in
\mathbb{N} \backslash \{0\}, c_i,d \in \mathbb{Z} \backslash \{0\},
\end{eqnarray}
has its rightmost pole at $N=1$ and is otherwise regular. In particular one may show that
\begin{eqnarray}
A_{\gamma e}^{(2)}(2) &=&
- \frac{64}{9} L^2
- 32 L
- \frac{149}{81}
+ \frac{736}{9} \zeta_2
- 128 \ln(2) \zeta_2 + 32 \zeta_3,
\\
A_{\gamma e}^{(2)}(3) &=&
- \frac{287}{72} L^2
- \frac{1961}{144} L
- \frac{870277}{10368}
+
\frac{1121}{18} \zeta_2,
\\
A_{\gamma e}^{(2)}(4) &=&
- \frac{869}{300} L^2
- \frac{87689}{9000} L
- \frac{10336457}{360000}
+ \frac{2027}{75} \zeta_2,
\end{eqnarray}
applying the algorithms of package {\tt HarmonicSums}
\cite{Vermaseren:1998uu,Blumlein:1998if,Ablinger:2014rba,Ablinger:2010kw, Ablinger:2013hcp, Ablinger:2011te,
Ablinger:2013cf,Ablinger:2014bra,Ablinger:2017Mellin}.
In $z$ space the OME is given by
\begin{eqnarray}
\label{eq:Ax}
      A_{\gamma e}^{(2)}(z) &=&
                \Biggl[
                        -\frac{16-28 z+11 z^2}{3 z}
                        +2 (2-z) {\rm H}_0
                        -\frac{4 \big(2-2 z+z^2\big)}{z}  {\rm H}_1
                \Biggr] L^2
                \nonumber \\ &&
                + \Biggl[
                        -\frac{2 \big(32-5 z+85 z^2\big)}{9 z}
                        +\frac{2 \big(32-32 z+31 z^2\big)}{3 z} {\rm H}_0
                        -2 (2-z) {\rm H}_0^2
                \nonumber \\ &&
                        +\frac{4 \big(20-8 z+13 z^2\big)}{3 z} {\rm H}_1
                        -\frac{12 \big(2-2 z+z^2\big)}{z} {\rm H}_1^2
                        +8 (2-z) {\rm H}_{0,1}
                \nonumber \\ &&
                        -8 (2-z) \zeta_2
                \Biggr] L
                +\frac{P_9}{135 z^3}
                -\frac{320-335 z+231 z^2}{15 z} {\rm H}_0
                +\frac{12+23 z}{6} {\rm H}_0^2
                +\frac{2-z}{3} {\rm H}_0^3
                \nonumber \\ &&
                +32 (2-z) \bigg(
                        \frac{(2-z)^2}{3 z^2}
                        - {\rm H}_0
                \bigg) \big( \tilde{\rm H}_{-1} \tilde{\rm H}_0 - \tilde{\rm H}_{0,-1} \big)
                -8 (2-z) {\rm H}_{0,0,1}
                \nonumber \\ &&
                -\frac{96-190 z+118 z^2-41 z^3}{3 z^2} {\rm H}_1^2
                -32 (2-z) \big(
                          \tilde{\rm H}_{-1} \tilde{\rm H}_ 0
                        - \tilde{\rm H}_{0,-1}
                \big) \tilde{\rm H}_1
                \nonumber \\ &&
                -\bigg(
                        \frac{2 \big(32-48 z+36 z^2-13 z^3\big)}{3 z^2}
                        +4 (2-z) {\rm H}_0
                \bigg) {\rm H}_{0,1}
                -\biggl(
                         \frac{2 P_{10}}{45 z^4}
                \nonumber \\ &&
                        -\frac{2 \big(32-48 z+12 z^2+7 z^3\big)}{3 z^2} {\rm H}_0
                \biggr) {\rm H}_1
                + \frac{ 2 \big(2-2 z+z^2\big)}{z}
                \biggl(
                        \frac{{\rm H}_1^3}{3}
                        + 8 {\rm H}_1 {\rm H}_{0,1}
                \nonumber \\ &&
                        + 16 \tilde{\rm H}_0 \tilde{\rm H}_{0,-1}
                        - 32 \tilde{\rm H}_{0,0,-1}
                        - 16 {\rm H}_{0,1,1}
                        + 8 \tilde{\rm H}_0 \zeta_2
                \biggr)
                +\biggl(
                         \frac{4 \big(32-48 z+24 z^2-3 z^3\big)}{3 z^2}
                \nonumber \\ &&
                        -8 (2-z) \big( {\rm H}_0 + 2 \tilde{\rm H}_1 \big)
                \biggr) \zeta_2
                +\frac{8 \big(12-10 z+5 z^2\big) }{z} \zeta_3,
\end{eqnarray}
with
\begin{eqnarray}
        P_9 &=& 1536-3072 z+1312 z^2-316 z^3-2005 z^4 ,
        \\
        P_{10} &=& 256-640 z-400 z^2+1320 z^3-1440 z^4+819 z^5 .
\end{eqnarray}
Here we refer to the harmonic polylogarithms \cite{Remiddi:1999ew},
\begin{eqnarray}
&&{\rm H}_{b,\vec{a}}(z) = \int_0^z dz f_b(z) {\rm H}_{\vec{a}}(z),~~{\rm H}_\emptyset = 1,~~f_0(z) = \frac{1}{z},~~f_1(z) =
\frac{1}{1-z},~~f_{-1}(z) = \frac{1}{1+z},
\nonumber\\
&& {\rm H}_{\underbrace{\mbox{\scriptsize 0 \ldots 0}}_k}(z) = \frac{1}{k!} \ln^k(z),
\end{eqnarray}
and define
\begin{eqnarray}
{{\rm H}}^*_{\vec{a}}(z) := {\rm H}_{\vec{a}}(1-z).
\end{eqnarray}

There are some terms in Eq.~(\ref{eq:Ax}) which are $\propto 1/z^4$. This is a reflection of terms $\propto
1/(N-4)$ in Eq.~(\ref{eq:AN}). One derives the small $z$ expansion of $A_{\gamma e}^{(2)}(z)$ which is
given by
\begin{eqnarray}
A_{\gamma e}^{(2)}(z) &=&
\biggl[
        - \frac{16}{3 z}
        +4/3
        + 4 {\rm H}_0
\biggr] L^2
+ \biggl[
        - \frac{64}{9z}
        + \frac{64}{3z} {\rm H}_0
        + \frac{250}{9}
        - 16 \zeta_2
        - \frac{64}{3} {\rm H}_0
        - 4 {\rm H}_0^2
\biggr] L
\nonumber\\ &&
+\frac{448}{27 z}
-\frac{796}{27}
+ 16 \zeta_3
+ (1 + 16 \zeta_2) {\rm H}_0 + 2 {\rm H}_0^2
 + \frac{2}{3} {\rm H}_0^3 + O(z),
\end{eqnarray}
showing that the most singular terms are of $O(1/z)$.

\section{The Radiators in \boldmath $z$ Space}
\label{sec:4}

\vspace*{1mm}
\noindent
We express the expansion coefficients (\ref{EQ:EXP1}--\ref{EQ:EXP2}) in terms of harmonic polylogarithms
\cite{Remiddi:1999ew}.
The corresponding expressions can be obtained using the package {\tt HarmonicSums} \cite{Vermaseren:1998uu,
Blumlein:1998if,Ablinger:2014rba,Ablinger:2010kw, Ablinger:2013hcp, Ablinger:2011te,Ablinger:2013cf,
Ablinger:2014bra,Ablinger:2017Mellin}, after reducing the algebraic relations \cite{Blumlein:2003gb}.
Because of the occurrence of some denominators $1/(1 \pm z)^l,~~l \in \mathbb{N},~~l > 1,$ one performs
the corresponding series expansion and uses summation techniques encoded in the packages {\tt Sigma}
\cite{SIG1,SIG2}, {\tt EvaluateMultiSums} and {\tt SumProduction} \cite{EMSSP}.
As usual, one has to separate the radiators into the part $\propto \delta(1-z)$, the contribution due
to $+$-distributions and the regular part,
\begin{eqnarray}
R(z,{{L}}) = R_\delta({{L}}) \delta(1-z)
+ [R_+(z,{{L}})]_+
+ R_{\rm reg}(z,{{L}}).
\end{eqnarray}
For the inclusive cross section the integral (\ref{eq:incl}) has to account for the fact that the
radiator $R(z,{{L}})$ is distribution--valued \cite{DISTR}.

For the inverse Mellin transform we use the notion {\tt PlusFunctionDefinition $\rightarrow$ 2} of the package {\tt
HarmonicSums},
\begin{eqnarray}
{\rm\bf M}\left[\frac{{\rm H}_{m_1,\vec{m}}(z) {\rm H}_1^k(z)}{1-z}\right](N) &=& \int_0^1 dz z^{N-1} \left[{\rm H}_{m_1,\vec{m}}(z) -
{\rm H}_{m_1,\vec{m}}(1) \right]
\frac{{\rm H}_1^k(z)}{1-z},~~~~m_1 \neq 1,
\\
{\rm\bf M}\left[\frac{{\rm H}_1^k(z)}{1-z}\right](N) &=& \int_0^1 dz \left(z^{N-1} - 1 \right)
\frac{{\rm H}_1^k(z)}{1-z}.
\end{eqnarray}
We use the following notation
\begin{eqnarray}
{\cal D}_k(z) = \left(\frac{\ln^k(1-z)}{1-z}\right)_+,~~~~k \in \mathbb{N},~~\text{with}~~{\rm H}_1(z) = - \ln(1-z).
\end{eqnarray}
The splitting function $P_{ee}^{(0)}$ is thus given by
\begin{eqnarray}
P_{ee}^{(0)}(z) &=&  8 {\cal D}_0(z) - 4 (1+z) + 6 \delta(1-z).
\end{eqnarray}
We use the following  conventions
\begin{eqnarray}
{\rm H}_{\vec{a}}(x) \equiv {\rm H}_{\vec{a}},~~~~~~~~~{\rm H}_{\vec{a}}(1-x) \equiv {\rm H}_{\vec{a}}^*,
\end{eqnarray}
The expressions up to $O(a^2)$ are given in Ref.~\cite{QED2019}. In the following we distinguish
the expansion coefficients in the vector and axial--vector case. Coefficients $c_{ij}$ without an index
$v$ or $a$ apply to both cases. We also display the difference terms
\begin{eqnarray}
\label{eq:DIFF}
c_{i,j}^{\rm \Delta} =
  c_{i,j}^{\rm a}
- c_{i,j}^{\rm v}.
\end{eqnarray}
The $\delta$--terms are
\input{xspaceD.tex}
\noindent
The $+$-distributions are given by
\input{xspaceP.tex}
\noindent
and the regular contributions read
\input{xspace1.tex}
The radiators depend on the polynomials $P_k|_{k=1}^{881}$ which are too voluminous to be displayed,
like also the radiators $c_{5,5}$ to $c_{6,5}$. They are given in an ancillary file to this paper.
The radiators exhibit evanescent poles $\propto 1/z^4$, which all cancel by performing an expansion
around $z = 0$ and the leading pole is again $1/z$.
\section{Numerical Results}
\label{sec:5}

\vspace*{1mm}
\noindent
In the following we study the effect of the radiators calculated on the $Z$ resonance. We extend previous work
\cite{Blumlein:2019pqb} to $O(\alpha^2)$ including the higher order corrections up to $O(\alpha^6 L^5)$ accounting
for the first three logarithmic corrections from $O(\alpha^3)$ to $O(\alpha^5)$.
\begin{figure}[H]
  \centering
  \includegraphics[width=0.6\linewidth]{figs/plot1.pdf}
  \caption{\sf
The ratio of the cross section containing all initial state corrections up
to $O(a)$ (dotted line),
to $O(a^2)$ (full line),
to $O(a^3 {L}^3)$ (dash-dotted line),
normalized on all terms including
also the $O(a^3 {L})$ and $O(a^4 {L}^4)$ corrections to the cross section $e^+ e^- \rightarrow \gamma^*/Z^*$
in
the region of the $Z$-peak
for $M_Z = 91.1876~\mathrm{GeV}$ \cite{PDG}.}
\label{FIG1}
\end{figure}
\begin{figure}[H]
  \centering
  \includegraphics[width=0.6\linewidth]{figs/plot2.pdf}
  \caption{\sf
The ratio of the cross section containing all initial state corrections  up
to $O(a^3 L^3)$ (dotted line),
to $O(a^3 L^2)$ (full line),
to $O(a^3 L)$ (dash-dotted line),
universal terms contributing to $O(a^3 L^0)$ (dashed line),
to $O(a^4 L^4)$ (long dashed line),
normalized on all terms including
also the $O(a^3 L)$ and $O(a^4 L^4)$ corrections to the cross section $e^+ e^- \rightarrow \gamma^*/Z^*$ in the region of the $Z$-peak
for $M_Z = 91.1876~\mathrm{GeV}$ \cite{PDG}.}
\label{FIG2}
\end{figure}
In Figures~\ref{FIG1} and \ref{FIG2} we compare the ratios of the three--loop terms with all contributions
(\ref{EQ:EXP1}--\ref{EQ:EXP2}) in the kinematic region  of $\sqrt{s} \in [85, 95]~\mathrm{GeV}$.

The $O(a)$ radiative corrections are large and amount to $\sim \pm 40\%$, followed by the $O(a^2)$ corrections, still
varying from $+15\%$ to $-7\%$, cf.~Figure~\ref{FIG1}. Already the leading term $O(a^3 {L}^3)$ yields only
corrections at the 1\% level.
The $O(a^3)$ corrections up to the $O(a^3 L)$ term are
significantly smaller and are illustrated in Figure~\ref{FIG2}.
\begin{table}[H]
    \centering
    \begin{tabular}{|l|r|r|r|r|}
    \hline
    \multicolumn{1}{|c|}{} &
    \multicolumn{2}{c|}{Fixed width} &
    \multicolumn{2}{c|}{$s$ dep. width} \\
    \hline
    \multicolumn{1}{|c|}{} &
    \multicolumn{1}{c|}{Peak} &
    \multicolumn{1}{c|}{Width}    &
    \multicolumn{1}{c|}{Peak} &
    \multicolumn{1}{c|}{Width} \\
    \multicolumn{1}{|c|}{} &
    \multicolumn{1}{c|}{(MeV)} &
    \multicolumn{1}{c|}{(MeV)}    &
    \multicolumn{1}{c|}{(MeV)} &
    \multicolumn{1}{c|}{(MeV)} \\
    \hline

    $O(\alpha)$   correction                 &   185.638   &   539.408  &   181.098  &   524.978 \\
    $O(\alpha^2 L^2)$:                       & -- 96.894   & --177.147  & -- 95.342  & --176.235 \\
    $O(\alpha^2 L)$:                         &     6.982   &    22.695  &     6.841  &    21.896 \\
    $O(\alpha^2 )$:                          &     0.176   & --  2.218  &     0.174  & --  2.001 \\
    $O(\alpha^3 L^3)$:                       &     23.265  &    38.560  &    22.968  &    38.081 \\
    $O(\alpha^3 L^2)$:                       & --   1.507  & --  1.888  & --  1.491  & --  1.881 \\
    $O(\alpha^3 L)$:                         & --   0.152  &     0.105  & --  0.151  & --  0.084 \\
    $O(\alpha^4 L^4)$:                       & --   1.857  &     0.206  & --  1.858  &     0.146 \\
    $O(\alpha^4 L^3)$:                       &      0.131  & --  0.071  &     0.132  & --  0.065 \\
    $O(\alpha^4 L^2)$:                       &      0.048  & --  0.001  &     0.048  &     0.001 \\
    $O(\alpha^5 L^5)$:                       &      0.142  & --  0.218  &     0.144  & --  0.212 \\
    $O(\alpha^5 L^4)$:                       & --   0.000  &     0.020  & --  0.001  &     0.020 \\
    $O(\alpha^5 L^3)$:                       & --   0.008  &     0.009  & --  0.008  &     0.008 \\
    $O(\alpha^6 L^6)$:                       & --   0.007  &     0.027  & --  0.007  &     0.027 \\
    $O(\alpha^6 L^5)$:                       & --   0.001  &     0.000  & --  0.001  &     0.000 \\

    \hline
    \end{tabular}
    \caption[]{\sf Shifts in the $Z$-mass and the width due to the different contributions to the ISR QED
    radiative corrections for a fixed width of $\Gamma_Z =  2.4952~\mathrm{GeV}$  and $s$-dependent width using
    $M_Z = 91.1876~\mathrm{GeV}$
    \cite{PDG} and $s_0 = 4 m_\tau^2$, cf.~\cite{ALEPH:2005ab}.\footnotemark}
    \label{TAB1}
    \end{table}
\footnotetext{In Ref.~\cite{Blumlein:2019pqb} slightly different numbers were reported for the shifts up to
the $O(a^2)$ corrections due to the use of $a(M_Z^2)$. Here we refer to $a(m_e^2)$.}

Finally, we summarize the shifts of the $Z$ peak and the corrections to the $Z$ width, $\Gamma_Z$, by the different
orders of the ISR radiative corrections in Table~\ref{TAB1}. Here we compare the results for the fixed and the
$s$-dependent width \cite{SDEP}. In Figures~\ref{FIG1} and \ref{FIG2} we depicted only the corrections up to
$O(a^3)$.
The other corrections up to $O(a^6 L^5)$ are only illustrated w.r.t. its shift of the $Z$ mass and change of the
$Z$ width since their behaviour is rather flat, yet they have an impact given the projected experimental resolutions.
The difference to the results in Ref.~\cite{Berends:1987ab} amounts to 4 MeV in
$\Gamma_Z$,~~\cite{Blumlein:2019pqb,QED2019}.

The relative shifts in adding the respective order can be positive or negative. at leading order $O(\alpha^k L^k)$
the level of $100~\mathrm{keV}$ \cite{DENT}\footnote{Statistical
accuracies of $\Delta M_Z = 0.005~\mathrm{MeV}~{\rm [stat]}, \Delta \Gamma_Z = 0.008~\mathrm{MeV}~{\rm [stat]}$ are quoted in
\cite{DENT}.} is only undershoot at $O(\alpha^5 L^4)$. Even the
$O(\alpha^4 L^2)$ corrections
are of the order of $50~\mathrm{keV}$, while at $O(\alpha^5 L^3)$ the level of $10~\mathrm{keV}$ is reached.
A missing link is still the $O(\alpha^3)$ term, which can be estimated to be roughly of $O(30~\mathrm{keV})$, setting the
frame of accuracy which is currently reached for the initial state corrections.
The QED corrections to 3--loop order are still somewhat larger than the experimental resolution at the the
FCC\_ee \cite{FCCEE} making the inclusion of also higher order subleading corrections necessary.

Furthermore, we remark that we have calculated the inclusive ISR corrections only, assuming that the
experimental data are extrapolated to the full phase space and only a cut in $s'$ is considered.
The experimental requirements may be more ambitious, requiring more differential radiative corrections in the
future. Due to different cuts, the corrections will turn out to be different and one has to carefully study
all the cut dependencies. For a recent summary see \cite{Jadach:2019huc}.
From the size of the corrections it seems that 3- to 4-loop corrections have there to
be provided too.

Numerical implementations for harmonic polylogarithms in {\tt Fortran} needed for the radiators in $z$ space are
given in
\cite{Gehrmann:2001pz} and \cite{Ablinger:2018sat}, respectively.
\section{Conclusions}
\label{sec:6}

\vspace*{1mm}
\noindent
We have calculated the QED initial state corrections to the annihilation process $e^+e^- \rightarrow
\gamma^*/Z^*$ up to the terms $O(\alpha^6 L^5)$. They come next to the recently completed $O(\alpha^2)$
corrections
\cite{Blumlein:2019srk,Blumlein:2019pqb,QED2019} and the well--known universal corrections $O((\alpha L)^k)$,
which were
known to fifth order in explicit form in the non--singlet and singlet case \cite{Skrzypek:1992vk,Jezabek:1992bx,
Przybycien:1992qe,Blumlein:1996yz,Arbuzov:1999cq,Arbuzov:1999uq,Blumlein:2004bs,Blumlein:2007kx}.
Here we included the first three logarithmic terms for the orders $O(\alpha^3)$ to $O(\alpha^5)$. The radiators
are given by convolutions of splitting functions, the contributions to the Wilson coefficient of the massless
Drell--Yan process and massive operator matrix elements. For the present corrections the massive OME
$\Gamma_{\gamma e}^{(1)}$ had to be calculated newly. The other massive OMEs were given in \cite{Blumlein:2011mi}
before. The corrections calculated in the present paper can be expressed in terms of harmonic
polylogarithms, if one also allows for the argument $(1-x)$ in case of harmonic polylogarithms containing an
index $i = -1$. In Mellin space the radiators can be represented in terms of harmonic sums and generalized
harmonic sums. The present corrections may still miss terms of $O(30~\mathrm{keV})$ for both $\delta M_Z$ and $\delta
\Gamma_Z$,  which can be further improved by  calculating the yet missing terms.

It is needless to say that by performing Mellin convolutions, using the quantities calculated in the present
paper and the massless quantities available in the literature, one is now in the position
to calculate all corrections of $O(\alpha^k L^k), O(\alpha^k L^{k-1})$ and $O(\alpha^k L^{k-2})$ for $k \geq 5$
straightforwardly. For the projected experimental accuracies in inclusive measurements at the FCC\_ee
they may not be needed beyond the orders already obtained.

redpajama_set_name: RedPajamaArXiv


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