# Integrating exp(x^2)

1. Sep 18, 2010

### Wishe Deom

Hello everyone,

This is my first post on the forum; I'm pretty sure my question fits in this section.

I am having a lot of difficulty finding the definite integral
$$\int^{+\infty}_{-\infty}e^{-2ax^{2}}$$ dx
where a is positive and real.
I know the answer is $$\\sqrt{\frac{\pi}{2a}}, but I have no idea at all how to get there. 2. Sep 18, 2010 ### fluxions hint: write another integral of that form, but with y as the variable. multiply them together. convert to polar coordinates, and integrate. then take the square root. 3. Sep 18, 2010 ### Wishe Deom I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products. Is this what you are suggesting I try? [tex]\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy 4. Sep 18, 2010 ### fluxions [tex] \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy$$

(most of the time)

and yes, i'm suggesting you look at:

$$\sqrt{\int \int e^{-\alpha x^2} e^{-\alpha y^2} dxdy$$

5. Sep 18, 2010

### fluxions

As a bonus, once you convince yourself that
$$\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{{\frac{\pi}{\alpha}}$$,
you can easily calculate integrals of the form
$$\int_{-\infty}^{\infty}x^{2n} e^{-\alpha x^2} dx$$
by differentiating the first expression (under the integral sign, on the left) with respect to alpha n times. This is a very useful fact.

6. Sep 18, 2010

### snipez90

To justify the interchange you can use "[URL [Broken] Theorem[/URL].

Last edited by a moderator: May 4, 2017
7. Sep 19, 2010

### Wishe Deom

Thank you for all the help. I'm well on my way, but I am having trouble with the limits of integration. Since x and y go from -inf to inf, what are the new limits?

Since r = sqrt(x^2+y^2), it seems that r goes from inf to inf, which would result in 0. I know, logically, r should be 0 to infinity, but how do I calculate that?

8. Sep 19, 2010

### HallsofIvy

This is "Fubini's Theorem"

9. Sep 19, 2010

### HallsofIvy

1) $$\int_{-\infty}^\infty e^{-2ax^2}dx= 2\int_0^\infty e^{-2ax^2}dx [/itex] by symmetry. 2) If you let [tex]I= \int_{-\infty}^\infty} e^{-2ax^2}dx$$
then $$\frac{I}{2}= \int_0^\infty e^{-2ax^2}dx$$
and it is also true that
$$\frac{I}{2}= \int_0^\infty e^{-2ay^2}dy$$
since that is the same integral with a different "dummy" variable.

so
3) $$I^2/4= \left(\int_0^\infty e^{-2ax^2}dx\right)\left(\int_0^\infty e^{-2ay^2}dy\right)$$
$$= \int_0^\infty \int_0^\infty e^{-2a(x^2+ y^2)} dy dx$$
by Fubini's theorem.

You can interpret that last as a double integral over the first quadrant and change to polar coordinates. Remember that the "differential of area" in polar coordinates is $r drd\theta$ and that $\theta$ goes from 0 to $\pi/2$ in the first quadrant.