# Trying to prove $\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+(\sin(\frac{\pi}{t}) -\frac{\pi}{t}\cos(\frac{\pi}{t}))^2}dt$

I posted this incorrectly several hours ago and now I'm back! So this time it's correct. Im trying to show that for $n\geq 1$:

$$\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}dt$$

I checked this numerically for several values of $n$ up through $n=500$ and the bounds are extremely tight.

I've been banging my head against this integral for a while now and I really can see no way to simplify it as is or to shave off a tiny amount to make it more palatable. Hopefully someone can help me. Thanks.

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Does anyone have any ideas?  I can't believe this thing is so difficult. – lithium barbie doll Jul 12 '12 at 2:12
What is the source? – Potato Jul 12 '12 at 2:14
Well Kokteil: unless there exists some witty trick, this seems to be an unbelievably hard inequality. I know you already said it is from do Carmo's book, but what page/number of example or exercise? And is it from "Differential Geometry of Curves and Surfaces"? – DonAntonio Jul 12 '12 at 2:15
Do Carmo's "Differential Geometry of Curves and Surfaces" Section 1-3 problem 9b, the integral is the arc length of a portion of the curve $(t,tsin(\frac{\pi}{t}))$. – lithium barbie doll Jul 12 '12 at 2:17
What it looks like is an elliptic integral (I know nothing about these, beyond apparently what they look like). – lithium barbie doll Jul 12 '12 at 2:20
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## 4 Answers

It is not hard to show (for example, in the problem following the problem you posted, exercise 10 in Section 1-3 of Differential Geometry of Curves and Surfaces by Do Carmo) that the arc length of and arc with endpoints $x$ and $y$ is at least the length of the straight line segment connecting them. In any case, the problem only asks for a "geometrical" proof.

That integral is the arc length of the curve $f(t) =(t,\sin (\pi/t))$ between the points $t=1/(n+1)$ and $t=1/n$. These points are $(1/(n+1), \sin((n+1)\pi)/(n+1))$ and $(1/n, \sin(n\pi)/n)$ (so the $y$ coordinates are $0$). Call them $A$ and $B$, respectively. The arc passes through the point $(1/(n+1/2),\sin(n\pi/2)/(n+(1/2))=(1/(n+1/2),\pm 1/(n+(1/2))$. Call this $C$. We see the arc length is at least the sum of the length of the segments $AC$ and $CB$. These each have length at least $\frac{1}{n+\frac{1}{2}}$ (draw the picture. This is the length of the perpendicular to the $x$-axis).

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Thank you for this! – lithium barbie doll Jul 12 '12 at 3:23

Potato's answer is what's going on geometrically. If you want it analytically:$$\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2} \geq \sqrt{\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}$$ $$= \bigg|\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\bigg|$$ The above expression is the absolute value of the derivative of $t\sin(\pi/t)$. So your integral is greater than $$\int_{1 \over n + 1}^{1 \over n + {1 \over 2}}|(t\sin({\pi \over t}))'|\,dt + \int_{1 \over n + {1 \over 2}}^{1 \over n}|(t\sin({\pi \over t}))'|\,dt$$ This is at least what you get when you put the absolute values on the outside, or $$\bigg|\int_{1 \over n + 1}^{1 \over n + {1 \over 2}}(t\sin({\pi \over t}))'\,dt\bigg| + \bigg|\int_{1 \over n + {1 \over 2}}^{1 \over n}(t\sin({\pi \over t}))'\,dt\bigg|$$ Then the fundamental theorem of calculus says this is equal to the following, for $f(t) = t \sin(\pi/t)$: $$\bigg|f({1 \over n + {1 \over 2}}) - f(0)\bigg| + \bigg|f({1 \over n}) - f({1 \over n + {1 \over 2}})\bigg|$$ $$= \bigg|{1 \over n + {1 \over 2}} - 0\bigg| + \bigg|0 -{1 \over n + {1 \over 2}}\bigg|$$ $$= {2 \over n + {1 \over 2}}$$

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I checked numerically that removing the 1 makes the inequality untrue, I don't think you can in general integrate the function $|\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)|$ without first pulling the absolute value bars outside the integral; consider $|x|$ in an interval around zero. – lithium barbie doll Jul 12 '12 at 3:25

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What I wrote was correct ($\int|f| \geq |\int f|$). I suggest you recheck your numerical work. – Zarrax Jul 12 '12 at 3:40
You're right, my mistake, apparently I should be more wary of Wolfram's margin of error. – lithium barbie doll Jul 12 '12 at 18:15

Now that I have finished this, I see that it is similar to the approach that Zarrax used, but it looks a bit simpler, so I will post it in addition.

Using the following facts $$\sqrt{x^2+1}\ge|x|\tag{1}$$ $$|x-y|\ge\mathrm{sgn}(x)(x-y)\tag{2}$$ with a change of variables $t\mapsto 1/t$, we get $$\begin{align} &\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}\mathrm{d}t\\ &=\int_n^{n+1}\sqrt{1+(\sin(\pi t)-\pi t\cos(\pi t))^2}\frac{\mathrm{d}t}{t^2}\\ &\ge\int_0^1\left|\frac{\pi \cos(\pi t)}{n+t}-\frac{\sin(\pi t)}{(n+t)^2}\right|\mathrm{d}t\\ &\ge\int_0^1\mathrm{sgn}(\cos(\pi t))\left(\frac{\pi \cos(\pi t)}{n+t}-\frac{\sin(\pi t)}{(n+t)^2}\right)\mathrm{d}t\\ &=\int_0^1\mathrm{sgn}(\cos(\pi t))\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\sin(\pi t)}{n+t}\right)\mathrm{d}t\\ &=\left(\frac{\sin(\pi/2)}{n+\frac12}-\frac{\sin(0)}{n}\right)+\left(\frac{\sin(\pi/2)}{n+\frac12}-\frac{\sin(\pi)}{n+1}\right)\\ &=\frac{2}{n+\frac12}\tag{3} \end{align}$$

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Hint: notice that the integral may be written as: $$\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}dt= \int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+{\left[\left(t\sin\left(\frac{\pi}{t}\right)\right)'\right]}^2}dt=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1 + [f(t)']^2}dt$$ where $f(t)= t \sin\frac{\pi}{t}.$ That means that you need to prove that the lenght of the graph of the function $f(t)$ from $\frac{1}{n+1}$ to $\frac{1}{n}$ is greater than or equal to $\frac{2}{n+\frac{1}{2}}$. What does it happen with the function $f(t)= t \sin\frac{\pi}{t}$ at the points $\frac{1}{n+1}$ and $\frac{1}{n}$? Geometrically, the result is evident.

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