# Thread: Integrating by Polar Substitution?

1. ## Integrating by Polar Substitution?

Hey,

How would you solve the following integral:

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

I've been trying to use a substitution, but I kind of forgot how to do it exactly. I've tried both polar and cylindrical coordinates, but I can't crunch it to the right answer. Any help? Sorry its been a while since I've done this type of integral.

2. Originally Posted by mbaboy
Hey,

How would you solve the following integral:

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

I've been trying to use a substitution, but I kind of forgot how to do it exactly. I've tried both polar and cylindrical coordinates, but I can't crunch it to the right answer. Any help? Sorry its been a while since I've done this type of integral.
Note that $\displaystyle e^{-\frac{x^2}{2}}$ is symmetric about the y axis, so $\displaystyle \int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}\,dx=2\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx$.

Now, let $\displaystyle I=\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx=\int_0^{\infty}e^{-\frac{y^2}{2}}\,dy$. Thus, $\displaystyle I^2=\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx\int_0^{\infty}e^{-\frac{y^2}{2}}\,dy=\int_0^{\infty}\int_0^{\infty}e ^{-\frac{1}{2}\left(x^2+y^2\right)}\,dy\,dx$

Now, make the conversion over to polar.

I leave you to verify that $\displaystyle \int_0^{\infty}\int_0^{\infty}e^{-\frac{1}{2}\left(x^2+y^2\right)}\,dy\,dx\implies\i nt_0^{\frac{\pi}{2}}\int_0^{\infty}e^{-\frac{r^2}{2}}r\,dr\,d\theta$

Now its simple integration from there.

In the end, you will end up with $\displaystyle I^2=\dots\implies I=\sqrt{\dots}$. But to get your final answer, multiply your value for $\displaystyle I$ by two.

Can you take it from here?

3. Thank you this is a very beautiful solution. I can now solve it algorithmically, but can you possibly provide any explanation to
$\displaystyle I=\int_0^{\infty}e^{-\frac{x^2}{2}}dx=\int_0^{\infty}e^{-\frac{y^2}{2}}dy$

I know it works, but I'm thinking that this implies $\displaystyle x=y$ which then wouldn't make much sense when converting into polar. Could you basically explain out the line where you introduce $\displaystyle I$?

4. Originally Posted by mbaboy
Thank you this is a very beautiful solution. I can now solve it algorithmically, but can you possibly provide any explanation to
$\displaystyle I=\int_0^{\infty}e^{-\frac{x^2}{2}}dx=\int_0^{\infty}e^{-\frac{y^2}{2}}dy$

I know it works, but I'm thinking that this implies $\displaystyle x=y$ which then wouldn't make much sense when converting into polar. Could you basically explain out the line where you introduce $\displaystyle I$?
I don't know how to call it in English... Actually, x is a variable you'll use only for the integral. It's just a name, it doesn't matter how it is called. You can call it y, z, a, b, whatever.
It just has to design the same thing in the integral.

The aim of writing I with x and with y is because we'll associate the two integrals.

5. Originally Posted by Moo
I don't know how to call it in English.
I think the english word is dummy variable.