A solution Integral of the Week #2 [Walking Randomly] « Cálculos auxiliares de problemas|teoremas

Julho 24, 2008

A solution Integral of the Week #2 [Walking Randomly]

Arquivado em: Uncategorized — Américo Tavares @ 9:20 am

I’ve just submitted the following solution for the

« Integral of the Week #2

The second Integral Of The Week (IOTW) is rather different from the first in that I am going to give you the evaluation. Your task is to prove it.

$\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=\sqrt{\pi}$

But WAIT! Almost every time I have seen this integral evaluated, it has been done by squaring it and converting to polar co-ordinates and that’s the one method of evaluation you can’t use for this particular challenge. I am looking for more ‘interesting’ proofs. Have fun.

Solutions can be posted in the comments section or sent to me by email (obtaining my email address is another puzzle for you to solve) and will be discussed in a future post. Feel free to send your solution in just about any format you like – plain text, uncompiled Latex, PDF, postscript, Mathematica, ODF, even Microsoft Word. When I get around to posting the solutions I will attempt to standardize them (to PDF probably).

By the way – you still have time to submit a solution for the first IOTW.

July 22nd, 2008 »

Solution

First I show that

$\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dx=I$ ,

where

$I=\displaystyle\int_{0}^{\infty}e^{-x^2}\; dx$ .

Then I evaluate $I^2$ to derive the value of $I$ by reversing the order of integration.

The second Integral Of The Week equals $2I$ :

$\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=2I$ .

If $x>0$ , then

$\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dx=x\displaystyle\int_{0}^{\infty}e^{-{xy}^2}\; dx$

$=x\displaystyle\int_{0}^{\infty}e^{-u^2}\dfrac{du}{x}=\displaystyle\int_{0}^{\infty}e^{-u^2}\; du=I$ .

Thus

$I^2=\displaystyle\int_{0}^{\infty}e^{-x^2}\; dx\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=\displaystyle\int_{0}^{\infty}dy\displaystyle\int_{0}^{\infty}xe^{-x^2}e^{-x^{2}y^{2}}\; dx$

$=\displaystyle\int_{0}^{\infty}dy\displaystyle\int_{0}^{\infty}xe^{-x^{2}(1+y^2)}\; dx$ $=\displaystyle\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty }$

$=\displaystyle\int_{0}^{\infty }\dfrac{-e^{-\underset{x\rightarrow \infty}{\lim }x}+e^{0}}{2\left( 1+y^{2}\right) }\; dy=\displaystyle\int_{0}^{\infty }\dfrac{1}{2\left( 1+y^{2}\right) }\; dy$