# An integral with variables

1. Oct 17, 2009

### forumfann

Could anyone help me evaluate the integral
$\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}|sx+ty|e^{-s^{2}/2}e^{-t^{2}/2}dsdt$, which should be a function of x and y?

By the way, this is not a homework problem.

Thanks

Last edited: Oct 17, 2009
2. Oct 17, 2009

### arildno

Well, make a shift to polar coordinates:
$$s=r\cos\theta,t=r\sin\theta$$
$$x=R\cos\phi,y=R\sin\phi$$

Thus, your integral becomes:
$$R\int_{0}^{\infty}\int_{0}^{2\pi}|\cos(\theta-\phi)|r^{2}e^{-\frac{r^{2}}{2}}d\theta{d}r$$

3. Oct 17, 2009

### forumfann

Thanks a lot for arildno's help. So I am able to get the value of the integral with 2 variable now.

But then how about 3 variables, i.e.
$$\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}|rx+sy+tz|e^{-r^{2}/2}e^{-s^{2}/2}e^{-t^{2}/2}drdsdt ?$$

4. Oct 17, 2009

### arildno

Spherical coordinates, perchance??

5. Oct 17, 2009

### Count Iblis

Instead of polar or spherical coordinates, you can also rotate your axis in the (r,s,t,...) space so that one of your axis becomes aligned with the (x,y,z,....) vector. The expression in the exponential is invariant under ratations, so what happens is that the integration becomes:

Integral dt1 dt2.....dtn |y t1| exp(-t1^2/2)exp(-t2^2/2)...
exp(-tn^2/2) =

2|y| (2pi)^[(n-1)/2]

where, of course, y = the length of your (x,y,z,...) vector

Last edited: Oct 17, 2009
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