Method NO. 3 : Neat Integral

16Sep08

Today, I decided to post a very eye-catching method for calculating an integral which shows very often in Statistical Mechanics in Physics.

Let’s try to calc this $\int\limits_{-\infty}^{\infty}e^{-x^{2}} dx$

First let’s mark it with a name $K$ , then

$K^{2} = \int\limits_{-\infty}^{\infty}e^{-x^{2}} dx \int\limits_{-\infty}^{\infty}e^{-y^{2}} dy = \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-(x^{2}+y^{2})} dx dy$

What is just a double integral over the whole 2d surface. It can be also written in polar coordinate system as $K^{2} = \int\limits_{0}^{\infty}e^{-r^{2}} \cdot 2\pi r dr = \pi \cdot \int\limits_{0}^{\infty}e^{-t} dt = \pi$

What means that $K = \int\limits_{-\infty}^{\infty}e^{-x^{2}} dx = \sqrt{\pi}$

Filed under: Mathematics, Methods | 7 Comments
Tags: coordinate, integral, math, method, pi, polar

7 Responses to “Method NO. 3 : Neat Integral”

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1 Osama Gamal on September 17, 2008 said:

I think we studied this method in the faculty 😀
Anyway, Well done 😉

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2 the7new7ramanujan on October 17, 2008 said:

i was also fascinated by it when i saw it for the first time.

Reply
3 rohedi on November 29, 2008 said:

maybe you need to give an explanation why the gaussian integral must be solved in polar cordinate, and also important to introduce the definition of gamma function.

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4 masteranza on November 29, 2008 said:

I don’t think I’ll have to explain that – it’s just another way to represent a given function in which it’s easier to integrate.
It’s just like using Cauchy instead Heine def of counting the limit.
I don’t have any clue why I should introduce the def of gamma function either.

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5 rohedi on November 29, 2008 said:

Sorry, here I want to inform you all that in my website on this address http://rohedi.com/content/view/34/1/, I post a new integral so-called Bernoulli Integral.
Maybe useful for you. Thanks.

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6 shah on December 22, 2008 said:

can u brief how u integrate the 2nd surface, can’t u?

Reply
7 masteranza on December 22, 2008 said:

Well, i’m integrating thin rings of radius r over all surface.

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