[SOLVED] Gaussian Integral with Polynomial Term

**powerlt** · March 29th 2010, 01:49 AM

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):

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

Apologies for a rather straightforward question and thanks in advance.

**Deadstar** · March 31st 2010, 05:52 AM

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...

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

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

**powerlt** · March 31st 2010, 02:17 PM

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...

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

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

**Deadstar** · March 31st 2010, 02:54 PM

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.

**simplependulum** · March 31st 2010, 08:33 PM

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

**powerlt** · April 1st 2010, 02:08 AM

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

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