# Thread:

1. ## [SOLVED] Gaussian Integral with Polynomial Term

Any help on how to calculate the following would be highly appreciated (I am familiar with the "standard" way of calculating the Gaussian integral in polar coordinates):

$%5Cint_%7B-%5Cinfty%20%7D%5E%7B%5Cinfty%7De%5E%7B-%5Cfrac%7B1%7D%7B2%7D%28x%5E%7B2%7D+%5Cfrac%7B1%7D%7Bx%5E2%7D%29%7Ddx$

Apologies for a rather straightforward question and thanks in advance.

2. YES I GOT IT!

Sorry... This has been boggling my mind for a wee while.

This integral involves the following lemma...

$\int_{-\infty}^{\infty} F(u) dx = \int_{-\infty}^{\infty} F(x) dx$ where $u = x-\frac{1}{x}$.

So we have...

$\int_{-\infty}^{\infty} e^{-\tfrac{1}{2}\left(x^2 + \tfrac{1}{x^2}\right)}dx$

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}\left(x - \tfrac{1}{x}\right)^2}dx$

WHY..?

Because $e^{-\tfrac{1}{2}\left(x-\tfrac{1}{x}\right)^2} = e^{-\tfrac{1}{2}\left(x^2 - 2 + \tfrac{1}{x^2}\right)} = e^{-\tfrac{1}{2}\left(x^2 + \tfrac{1}{x^2}\right)}e$

So we need an extra $e^{-1}$ outside the integral!

So continuing we get...

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}\left(x - \tfrac{1}{x} \right)^2}dx$

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}x^2}dx$ using Lemma!

$=\frac{1}{e} \sqrt{2} \sqrt{\pi}$

$= \frac{\sqrt{2\pi}}{e}$

Epic Win!

Additional calculations

Showing that

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}x^2}dx = \frac{1}{e} \sqrt{2} \sqrt{\pi}$

Let $z = \frac{x}{\sqrt{2}}$, then $dx = \sqrt{2}dz$.

Hence the integral becomes...

$\frac{\sqrt{2}}{e} \int_{-\infty}^{\infty} e^{-z^2}dz$

$=\frac{\sqrt{2 \pi}}{e}$

3. Thanks, Deadstar. Any chance you could expand a bit/point me towards the proof of the lemma?

Originally Posted by Deadstar
YES I GOT IT!

Sorry... This has been boggling my mind for a wee while.

This integral involves the following lemma...

$\int_{-\infty}^{\infty} F(u) dx = \int_{-\infty}^{\infty} F(x) dx$ where $u = x-\frac{1}{x}$.

So we have...

$\int_{-\infty}^{\infty} e^{-\tfrac{1}{2}\left(x^2 + \tfrac{1}{x^2}\right)}dx$

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}\left(x - \tfrac{1}{x}\right)^2}dx$

WHY..?

Because $e^{-\tfrac{1}{2}\left(x-\tfrac{1}{x}\right)^2} = e^{-\tfrac{1}{2}\left(x^2 - 2 + \tfrac{1}{x^2}\right)} = e^{-\tfrac{1}{2}\left(x^2 + \tfrac{1}{x^2}\right)}e$

So we need an extra $e^{-1}$ outside the integral!

So continuing we get...

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}\left(x - \tfrac{1}{x} \right)^2}dx$

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}x^2}dx$ using Lemma!

$=\frac{1}{e} \sqrt{2} \sqrt{\pi}$

$= \frac{\sqrt{2\pi}}{e}$

Epic Win!

Additional calculations

Showing that

$=\frac{1}{e} \int_{-\infty}^{\infty} e^{-\tfrac{1}{2}x^2}dx = \frac{1}{e} \sqrt{2} \sqrt{\pi}$

Let $z = \frac{x}{\sqrt{2}}$, then $dx = \sqrt{2}dz$.

Hence the integral becomes...

$\frac{\sqrt{2}}{e} \int_{-\infty}^{\infty} e^{-z^2}dz$

$=\frac{\sqrt{2 \pi}}{e}$

4. EDIT: HAHAHA I just realized you were after the proof, not an explanation of how I used it... I'll post it up next post, I don't understand it fully though...

Sure. Bare in mind I just learned this Lemma so I hope I'm describing it correctly!

So, the idea is, if we have an integral of the form $\int_{-\infty}^{\infty} F(u) dx$ where $u=x - \frac{1}{x}$

This is equal to the integral $\int_{-\infty}^{\infty} F(x) dx$.

Where we replace the expression $x - \frac{1}{x}$ by just x.

So an example... (I think)

$\int_{-\infty}^{\infty} \frac{1}{x - \frac{1}{x}} dx = \int_{-\infty}^{\infty} \frac{1}{x} dx$.

As by the lemma... We can replace the expression $x - \frac{1}{x}$ by just $x$.

So... In the example you posted... I just rewrote $x^2 + \frac{1}{x^2}$ as $\left(x - \frac{1}{x}\right)^2$ and included an $e^{-1}$ outside the integral to cancel out the extra $e$ created.

So...

$e^{-\frac{1}{2}\left(x^2 + \frac{1}{x^2}\right)}$

$= \frac{e}{e}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2x^2}}$

$= \frac{1}{e}e^{-\frac{1}{2}x^2} e^1 e^{-\frac{1}{2x^2}}$

$= \frac{1}{e}e^{-\frac{1}{2}x^2 + 1 - \frac{1}{2x^2}}$

$= \frac{1}{e}e^{-\frac{1}{2}\left(x^2 - 2 + \frac{1}{x^2}\right)}$

$= \frac{1}{e}e^{-\frac{1}{2}\left(x - \frac{1}{x}\right)^2}$

However, I did it the opposite way round and just multiplied the integral by the inverse of the extra term I had.

But anyway... So we have the integral

$\int_{-\infty}^{\infty} \frac{1}{e}e^{-\frac{1}{2} \left(x - \frac{1}{x}\right)^2} = \frac{1}{e}\int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(x - \frac{1}{x}\right)^2}$

and by the Lemma, we can replace the $x - \frac{1}{x}$ term by $x$.

Hence we are left with...

$\frac{1}{e} \int_{-\infty}^{\infty} e^{-\frac{1}{2}x^2}$ which is much more workable as shown in my additional calculations.

This clear things up? I'm still learning about it as I say so hopefully one of the mods could step in and say whether I'm doing things right.

5. Almost a year ago , someone posted it as a challenge . I think this thread has been passed into oblivion ...

http://www.mathhelpforum.com/math-he...ation-1-a.html

6. Thanks for the link, simplependulum!
And in general - really valuable input guys - appreciate it.
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