# Integral of exp function?

1. Dec 22, 2008

### philip041

I am meant to get the integral

$$\int^{\infty}_{-\infty}{dxe^{-\alpha^2x^2}}$$

to equal

$$A^2\frac{\sqrt{\pi}}{\alpha}$$

but I get

$$A^2\left[\frac{e^{-\alpha^2x^2}}{\alpha^2}\right]^{\infty}_{-\infty}$$

which I'm pretty sure doesn't equal much.

Thoughts?

Last edited by a moderator: Dec 22, 2008
2. Dec 22, 2008

### philip041

Argh I don't know how to sort the Latex out, suggest how and I will edit it, or if you want repost my stuff. It works in my Latex I don;t know what I'm doing wrong!

3. Dec 22, 2008

### dirk_mec1

The slash should be the other way around like this:

$$\int^{\infty}_{-\infty}{e^{-\alpha^2x^2}} \mbox{d}x$$

4. Dec 22, 2008

### Staff: Mentor

Your closing tex tags use a backslash but should use a forward slash. I have changed them in what I've quoted from you.

Regarding the integral, you don't show how you got what you show. Inasmuch as this is an improper integral, you probably should have split this into two integrals and used limits.

Also, you seem to have gotten $$e^{-a^2x^2}/a^2$$ as your antiderivative, which is incorrect. To see that this is so, take the derivative.

5. Dec 22, 2008

### HallsofIvy

Staff Emeritus
That's a fairly standard calculation but because the anti-derivative is NOT an "elementary" function, requires a couple of tricks to do that definite integral.

You want to find
$$I= \int_{-\infty}^{\infty} e^{-\alpha^2x^2}dx$$

First, because the integrand is symmetric about x= 0,
$$\frac{I}{2}= \int_0^\infty e^{-\alpha^2x^2}dx[/itex] It is also true that [tex]\frac{I}{2}= \int_0^\infty e^{-\alpha^2y^2}dy[/itex] so that [tex]\frac{I^2}{4}= \left(\int_0^\infty e^{-\alpha^2x^2}dx\right)\left(\int_0^\infty e^{-\alpha^2y^2}dy\right)$$
$$= \int_{R^+\times R^+}\int e^{-\alpha^2(x^2+ y^2)}dA$$
Where the integral is over the first quadrant. Write that in polar coordinates and it becomes easy to integrate.

6. Dec 22, 2008

Cheers