# Evaluating a double integral to find its signle integral.

• May 17th 2008, 11:31 PM
flawless
Evaluating a double integral to find its signle integral.
Evaluate I=integral[e^(-x^2)dx] from 0 to infinity,
by computing I^2= double integral[[e^(-x^2)e^(-y^2)dxdy]] both integrals from 0 to infinity. (hint: use appropriate coordinates (i think they mean maybe polar??))

(hope you can understand the math i wrote) Thanks
• May 17th 2008, 11:41 PM
flyingsquirrel
Hi
Quote:

Originally Posted by flawless
i think they mean maybe polar??

Yes, the polar coordinates $(r,\,\theta)$ are the appropriate ones. What can you get from it ?
• May 17th 2008, 11:44 PM
Moo
Hello,

Quote:

Originally Posted by flawless
Evaluate I=integral[e^(-x^2)dx] from 0 to infinity,
by computing I^2= double integral[[e^(-x^2)e^(-y^2)dxdy]] both integrals from 0 to infinity. (hint: use appropriate coordinates (i think they mean maybe polar??))

(hope you can understand the math i wrote) Thanks

Yes, it's polar coordinates.

$x=r \cos \theta$
$y=r \sin \theta$

Therefore, $dxdy$ becomes $r dr d\theta$ (I know it is the determinant of the Jacobian matrix, but I don't know how you call it).

--> $I^2=\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy=\int_0^{\frac{\pi}{2}} \int_0^\infty r e^{-r^2} dr d \theta=\frac{\pi}{2} \cdot \frac{1}{2}=\frac{\pi}{4}$

$\implies I=\frac{\sqrt{\pi}}{2}$
• May 17th 2008, 11:49 PM
flyingsquirrel
Quote:

Originally Posted by Moo
--> $I^2=\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy=\int_0^{{\color{red}2 \pi}} \int_0^\infty r e^{-r^2} dr d \theta=2 \pi \cdot \frac{1}{2}=\pi$

$x$ and $y$ are positive hence $\theta$ lies in $\left[0,\frac{\pi}{2}\right]$.
• May 18th 2008, 12:01 AM
Moo
I forgot it (Bow)
• May 18th 2008, 12:03 AM
wingless
$\int_{0}^{\infty}e^{-x^2}~dx$

Let's write it two times.

$\int_{0}^{\infty}e^{-x^2}~dx~ \int_{0}^{\infty}e^{-x^2}~dx$

Let x be y on the second integral.

$\int_{0}^{\infty}e^{-x^2}~dx~ \int_{0}^{\infty}e^{-y^2}~dy$

By the properties of double integral:

$\int_{0}^{\infty}\int_{0}^{\infty}e^{-x^2-y^2}~dx~dy$

Let $x = r\cos \theta$ and $y = r\sin \theta$.
Then $-x^2-y^2 = -r^2$ and $dx~dy = r~dr~d\theta$.

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-r^2}r~dr~d\theta$

$\int_{0}^{\frac{\pi}{2}}~d\theta~\int_{0}^{\infty} r e^{-r^2}~dr$

$\frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$

So,

$\int_{0}^{\infty}e^{-x^2}~dx~ \int_{0}^{\infty}e^{-x^2}~dx = \frac{\pi}{4}$

$\int_{0}^{\infty}e^{-x^2}~dx = \frac{\sqrt{\pi}}{2}$
• May 18th 2008, 06:37 AM
Krizalid
Let $0 Start by showing that $\Gamma (z)\Gamma (1-z)=\int_{0}^{\infty }{\frac{x^{z-1}}{1+x}\,dx}.$ Set $z=\frac12\implies\Gamma \left( \frac{1}{2} \right)^{2}=\pi \,\therefore \,\Gamma \left( \frac{1}{2} \right)=\sqrt{\pi }.$ Finally $\int_{0}^{\infty }{e^{-x^{2}}\,dx}=\frac{1}{2}\Gamma \left( \frac{1}{2} \right)=\frac{\sqrt{\pi }}{2}$ as required. $\blacksquare$

(There're may proofs anyway.)