View Full Version : An integral with variables
forumfann
Oct17-09, 12:38 PM
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
arildno
Oct17-09, 01:00 PM
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
forumfann
Oct17-09, 03:01 PM
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 ?
arildno
Oct17-09, 03:03 PM
Spherical coordinates, perchance??
Count Iblis
Oct17-09, 05:43 PM
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
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