# Thread: Evaluating a double integral to find its signle integral.

1. ## 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

2. Hi
Originally Posted by flawless
i think they mean maybe polar??
Yes, the polar coordinates $\displaystyle (r,\,\theta)$ are the appropriate ones. What can you get from it ?

3. Hello,

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.

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

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

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

$\displaystyle \implies I=\frac{\sqrt{\pi}}{2}$

4. Originally Posted by Moo
--> $\displaystyle I^2=\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy=\int_0^{{\color{red}2 \pi}} \int_0^\infty r e^{-r^2} dr d \theta=2 \pi \cdot \frac{1}{2}=\pi$
$\displaystyle x$ and $\displaystyle y$ are positive hence $\displaystyle \theta$ lies in $\displaystyle \left[0,\frac{\pi}{2}\right]$.

5. I forgot it

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

Let's write it two times.

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

Let x be y on the second integral.

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

By the properties of double integral:

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

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

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

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

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

So,

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

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

7. Let $\displaystyle 0<z<1.$ Start by showing that $\displaystyle \Gamma (z)\Gamma (1-z)=\int_{0}^{\infty }{\frac{x^{z-1}}{1+x}\,dx}.$ Set $\displaystyle z=\frac12\implies\Gamma \left( \frac{1}{2} \right)^{2}=\pi \,\therefore \,\Gamma \left( \frac{1}{2} \right)=\sqrt{\pi }.$ Finally $\displaystyle \int_{0}^{\infty }{e^{-x^{2}}\,dx}=\frac{1}{2}\Gamma \left( \frac{1}{2} \right)=\frac{\sqrt{\pi }}{2}$ as required. $\displaystyle \blacksquare$

(There're may proofs anyway.)