# Compute the following integral?

Posted 7 months ago
1030 Views
|
3 Replies
|
1 Total Likes
|
 I let Mathematica run overnight the following code Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity}, {x, 0, Pi/2}] but it was unable to compute the closed form in that time. However, Mathematica also didn’t “give up”. Do you think Mathematica can do it?
3 Replies
Sort By:
Posted 7 months ago
 I suspect that the Cos inside a Cos makes it impossible. It looks like an integral in polar coordinates. Might be easier in Cartesian coordinates.
 Doing the r under assumptions is the first step In[2]:= Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity},(* {x,0,Pi/2} *) Assumptions -> {A > 0, a > 0, b > 0}] Out[2]= (E^(-(1/2) A^2 Cos[x]^2) Sqrt[\[Pi]/2] (1 - A^2 Cos[x]^2))/( a^2 Cos[x]^2 + b^2 Sin[x]^2) now look for a meaningful substitution, you need $0 \leq x \leq \frac{\pi}{2}$. And as a preliminary step try $a = b$ to get rid of the denominator.