# Show that $\int\limits_{-\infty}^{\infty} \! \mathbb{e}^{-x^2} \, \mathrm{d}x$ = $\sqrt{\pi}$. [duplicate]

Possible Duplicate:
Explain $\iint \mathrm dx\mathrm dy = \iint r \mathrm d\alpha\mathrm dr$

Show that $\int\limits_{-\infty}^{\infty} \! \mathbb{e}^{-x^2} \, \mathrm{d}x$ = $\sqrt{\pi}$.

The hint was to calculate the double integral $\int\limits_{-\infty}^{\infty} \! \int\limits_{-\infty}^{\infty} \! \mathbb{e}^{-x^2}\mathbb{e}^{-y^2} \, \mathrm{d}x\mathrm{d}y$ using polar coordinates. The double integral equals [$\int\limits_{-\infty}^{\infty} \! \mathbb{e}^{-x^2} \, \mathrm{d}x]^2$.

So I started by writing $x=r\cos(\theta)$ and $y=r\sin(\theta)$. I know that $x^2+y^2=r^2$. Then I need to change those variables in my original equation, but this confuses me.

I get:

$$\int\limits_{-\infty}^\infty \! \int\limits_{-\infty}^\infty \! \mathbb{e}^{-r^2}\, \mathrm{d}x\,\mathrm{d}y$$

I also got told that the double integral equals:

$$\int\limits_0^\infty \! \int\limits_{0}^{2\pi} \! \mathbb{e}^{-r^2}r \, \mathrm{d}\theta\,\mathrm{d}r$$

Why though?

Thanks for any help in advance.

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## marked as duplicate by Did, Andrés Caicedo, Marvis, Lukas Geyer, Martin ArgeramiNov 30 '12 at 2:12

The extra factor $r$ comes from the Jacobian of the change of variables. If you're uncertain, check your textbook, or (for example) en.wikipedia.org/wiki/Multiple_integral#Change_of_variables – mrf Nov 29 '12 at 22:57
Are you asking why the double integral in Cartesian coordinates is equal to the double integral in polar coordinates? – Michael Hardy Nov 29 '12 at 22:58
@RossMillikan I don't think this is a duplicate since here the OP wants to know how $dx dy$ transformed into $r d r d \theta$. – user17762 Nov 29 '12 at 22:58
Indeed, I want to know how to change dxdy into rdrdθ, and how to change my interval of integration. – Jeroen Nov 29 '12 at 23:12
This was mistakenly closed as a duplicate. But the content of this question is: Why does $dx\,dy$ become $r\,dr\,d\theta$? The other question cited as an exact duplicate does not even mention that? – Michael Hardy Nov 29 '12 at 23:41

On the circle of radius $r$ centered at the origin, if the angle $\theta$ is incremented by an infinitely small amount $d\theta$, then the infinitely small arc length involved is $r\,d\theta$. An infinitely small increment $dr$ in $r$ is in a direction a right angles to the circle. So you're looking at an infinitely small rectangle whose sides have lengths $dr$ and $r\,d\theta$, and whose area is therefore $r\,dr\,d\theta$.

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