# Thread:

1. ## Another nasty integral

I've got another one for you.

I ran across this theorem in my Field Theory text. The theorem is this:

Given a sufficiently smooth function f(x) we have that:
$\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}$

The term "sufficiently smooth" does not require that f(x) is a $C^{\infty}$ function, but as most wavefunctions in Physics are $C^{\infty}$ I would be satisfied with a proof of this case. (Note: Obviously we are concerned only with f(x) that have limits that existat $\pm \infty$.)

The only f(x) I can figure out how to integrate (as a test of the theorem) is a Gaussian. All the other functions I know of with the required properties at infinity is discontinous at at least one point on the real line, so it doesn't qualify.

-Dan

2. Originally Posted by topsquark
I've got another one for you.

I ran across this theorem in my Field Theory text. The theorem is this:
What does this have to do with Field theory?

3. Originally Posted by topsquark
I've got another one for you.

I ran across this theorem in my Field Theory text. The theorem is this:

Given a sufficiently smooth function f(x) we have that:
limit(x -> + infinity)f(x) + limit(x -> - infinity) = limit(e -> 0+) e Integral( dx f(x) exp(-e*abs(x)) ) where the integral is over the whole real line.

The term "sufficiently smooth" does not require that f(x) is a C(infinity) function, but as most wavefunctions in Physics are C(infinity) I would be satisfied with a proof of this case. (Note: Obviously we are concerned only with f(x) that have limits that exist at (+/-) infinity.)

The only f(x) I can figure out how to integrate (as a test of the theorem) is a Gaussian. All the other functions I know of with the required properties at infinity is discontinous at at least one point on the real line, so it doesn't qualify.

-Dan
This is difficult to read without LaTeX, but this might be a consequence
of the Laplace transform derivative properties.

RonL

4. Originally Posted by ThePerfectHacker
What does this have to do with Field theory?
It comes in handy when deriving the interaction matrix (S matrix) using the path integral method. In the long run we are going to take the natural log of a quantity that is the integral of this, so it is convenient to change all factors into exponential terms. That's as much as I can give you without quoting some seriously nasty integrals. (Which I'm not about to do without LaTeX!)

-Dan

5. Originally Posted by CaptainBlack
This is difficult to read without LaTeX, but this might be a consequence
of the Laplace transform derivative properties.

RonL
Interesting thought. I'll look into it. Thanks!

-Dan

6. Still working on the clue, but since LaTeX is back I thought I'd fix up the original post in case it gives any thoughts.

-Dan

7. Pffl. I'm running into a problem almost immediately.
$\int_{-\infty}^{\infty}dx F(x) e^{-\epsilon |x|} = \int_0^{\infty}dx F(-x) e^{-\epsilon x} + \int_0^{\infty}dx F(x) e^{-\epsilon x}$

Although many wavefunctions are parity eigenstates I cannot assume this, so I have no way to simplify the first integral in general. My Laplace transforms abilities are introductory. Is there a formula in terms of $L \{F(x) \}$ for the first integral?

If I DO assume that F(x) is even/odd then I get:
$\int_{-\infty}^{\infty}dx F(x) e^{-\epsilon |x|} = \pm f(\epsilon) + f(\epsilon)$

So when I take $\lim_{\epsilon \to 0^+} \epsilon (\pm f(\epsilon) + f(\epsilon))$ I get nothing that resembles $F(\infty) + F(-\infty)$, it's in terms of the transform not the original function?

-Dan

8. I was thinking about this approach also. I tend to make things more complicated than they need to be so I thought I'd simply try an integration by parts:
$\int_{-\infty}^{\infty}dx f(x)e^{-\epsilon |x|} = - \lim_{x \to \infty} f(x) \frac{e^{-\epsilon x}}{\epsilon} - \lim_{x \to -\infty} f(x) \frac{e^{\epsilon x}}{\epsilon} -$ $\int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}$

$= - \lim_{x \to \infty} \left [ (f(x) + f(-x)) \frac{e^{-\epsilon x}}{\epsilon} \right ] - \int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}$

To finish this I need to operate $\lim_{\epsilon \to 0^+} \epsilon$ on this, so I get:
$= - \lim_{\epsilon \to 0^+} \epsilon \lim_{x \to \infty} \left [ (f(x) + f(-x)) \frac{e^{-\epsilon x}}{\epsilon} \right ] -$ $\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}$

$= - \lim_{\epsilon \to 0^+} \lim_{x \to \infty} \left [ (f(x) + f(-x)) e^{-\epsilon x} \right ] - 0?$

But how does one take this limit?

(I'm also having a problem arguing the limit of the integral, but I'll worry about that later.)

-Dan

9. Originally Posted by topsquark
Pffl. I'm running into a problem almost immediately.
$\int_{-\infty}^{\infty}dx F(x) e^{-\epsilon |x|} = \int_0^{\infty}dx F(-x) e^{-\epsilon x} + \int_0^{\infty}dx F(x) e^{-\epsilon x}$
I am sure you know this, you cannot take a Laplace transform of any given function. If a function is "too large" then it will not work. For example $e^{x^2}$.
I think the condition is (peicewise smooth),
$|f(x)|<K\cdot e^{-\epsilon x}$.
Futhermore, should it not be,
$\int_{-\infty}^0$
You have it the other way around.

10. Originally Posted by ThePerfectHacker
I am sure you know this, you cannot take a Laplace transform of any given function. If a function is "too large" then it will not work. For example $e^{x^2}$.
I think the condition is (peicewise smooth),
$|f(x)|<K\cdot e^{-\epsilon x}$.
Futhermore, should it not be,
$\int_{-\infty}^0$
You have it the other way around.
Actually, no. I subbed $x' = -x$ into the $\int_{-\infty}^0$ integral to get it into the form I posted.

So
$\int_{-\infty}^0 dx F(x) e^{\epsilon x} = \int_{\infty}^0 (-dx') F(-x') e^{-\epsilon x'}$ $= \int_0^{\infty} dx' F(-x') e^{-\epsilon x'}$
(I did this to put it into the "correct" form for a Laplace transform.)

As to the "largeness" argument, I am assuming a function that approaches a constant value at either infinity (ie I am assuming $\lim_{x \to \pm \infty}f(x)$ exists) and since I'm assuming f(x) is a $C^{\infty}$ function I am also assuming no discontinuities. (Perhaps I wasn't clear enough about that in my first post.) I believe (please correct me if I'm wrong) that guarentees the existence of the integral. If not then I am simply forced to add the assumption that the integral does exist and I'll work out later what restrictions that puts on the Physics.

-Dan

11. Originally Posted by topsquark
Actually, no. I subbed $x' = -x$ into the $\int_{-\infty}^0$ integral to get it into the form I posted.

As to the "largeness" argument, I am assuming a function that approaches a constant value at either infinity (ie I am assuming $\lim_{x \to \pm \infty}f(x)$ exists) and since I'm assuming f(x) is a $C^{\infty}$ function I am also assuming no discontinuities. (Perhaps I wasn't clear enough about that in my first post.) I believe (please correct me if I'm wrong) that guarentees the existence of the integral. If not then I am simply forced to add the assumption that the integral does exist and I'll work out later what restrictions that puts on the Physics.

-Dan
I would have a look at this, but its approaching Christmas, so I am run
off my feet here with getting stuff and attending end of term shows etc.

It may be a week before I get time in anything more than 10 minute slots.

RonL

12. Originally Posted by CaptainBlack
I would have a look at this, but its approaching Christmas, so I am run
off my feet here with getting stuff and attending end of term shows etc.

It may be a week before I get time in anything more than 10 minute slots.

RonL
Understood. I am going on a two week stint in Myrtle Beach with my parents (they're staying there until April) so I might not be around much myself.

(Chuckles) Don't worry. I suspect the problem will still be around when you get back. (Unless someone has a flash of inspiration anyway. But as I'll be away from my personal "library" it won't likely be me! )

-Dan

13. I got a tip from another forum on this. I want to bounce it off you guys.

For starters this is NOT rigorous, however it does explain why the assumption of a "sufficiently smooth" function was made as opposed to a more precise statement.

$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|}$

Split the integral into two parts, x < 0 and x > 0. I am going to look at the latter integral.
$\lim_{\epsilon \to 0^+} \epsilon \int_0^{\infty}dx f(x) e^{-\epsilon x}$

(*) Now make the substitution $y = \epsilon x$:
$\lim_{\epsilon \to 0^+} \epsilon \int_0^{\infty}dx f(x) e^{-\epsilon x} = \lim_{\epsilon \to 0^+} \int_0^{\infty}dy f \left ( \frac{y}{\epsilon} \right ) e^{-y}$

If the function is "sufficiently smooth" we may interchange the limit and integration operations:
$\lim_{\epsilon \to 0^+} \int_0^{\infty}dy f \left ( \frac{y}{\epsilon} \right ) e^{-y} = \int_0^{\infty}dy e^{-y} \lim_{\epsilon \to 0^+} f \left ( \frac{y}{\epsilon} \right )$

Now, $\lim_{\epsilon \to 0^+} f \left ( \frac{y}{\epsilon} \right )$ if it exists is just the number $\lim_{x \to \infty}f(x)$ and is a constant, so we may take it outside of the integration.
$\int_0^{\infty}dy e^{-y} \lim_{\epsilon \to 0^+} f \left ( \frac{y}{\epsilon} \right ) = f(\infty) \int_0^{\infty}dy e^{-y} = f(\infty) \cdot 1 = f(\infty)$
where I have used an obvious notation.

The x < 0 integration may be done in the same way. Thus we have:
$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|} = f(\infty) + f(-\infty)$

One comment. Go back to the (*) line. Being more careful, this integral should be written as:
$\lim_{\epsilon \to 0^+} \epsilon \lim_{N \to \infty} \int_0^{N}dx f(x) e^{-\epsilon x}$

So when I make the y substitution:
$\lim_{\epsilon \to 0^+} \epsilon \lim_{N \to \infty} \int_0^{N}dx f(x) e^{-\epsilon x} = \lim_{\epsilon \to 0^+} \lim_{N \to \infty} \int_0^{\epsilon N}dy f \left ( \frac{y}{\epsilon} \right ) e^{-y}$

In order to get the expression I wrote I implicitly took $\lim_{\epsilon \to 0^+} \lim_{N \to \infty} \epsilon N \to \infty$. The argument is that $\epsilon$, however small, is always a positive number so upon taking the N limit $\epsilon N$ should still go to infinity. This, aside from switching the limit and the integration, is the weakest point in the argument. Any comments?

Thanks!
-Dan

14. Originally Posted by topsquark
I've got another one for you.

I ran across this theorem in my Field Theory text. The theorem is this:

Given a sufficiently smooth function f(x) we have that:
$\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}$
Rough outline:

1. Every function can be written as the sum of an odd function and an even function zero at the origin and a constant.

2. The result is trivially true for a sufficiently well behaved odd function.

3. The result is a result of the fundamental theorem of calculus, and the
Laplace transform of a derivative for an even function (sufficiently well behaved) zero at the origin.

3a. The result is true for a constant (we need to use the Laplace transform of a constant result here)

4. Hence the result is true for a sufficiently well behaved function.

I can amplify the details if you need (though I doubt that I will feel like
going for full rigour)

RonL

15. Originally Posted by CaptainBlack
Rough outline:

1. Every function can be written as the sum of an odd function and an even function.

2. The result is trivially true for a sufficiently well behaved odd function.

3. The result is a result of the fundamental theorem of calculus, and the
Laplace transform of a derivative for an even function (sufficiently well behaved).

4. Hence the result is true for a sufficiently well behaved function.

I can amplify the details if you need (though I doubt that I will feel like
going for full rigour)

RonL
No, I don't need "full rigor." (Sounds dangerous anyway! ) I just wanted to check that there were no glaring logical arguments in the proof. For some reason when I integrate over arbitrary functions my intuition and confidence goes right out the window. Thanks for the review!

-Dan
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