Evaluating an integral over inifinty with polars leads to an integral of cosine over inifinity, how can this be resolved?

Question

So I have the integral $$\int_0^\infty\int_0^\infty\frac{yx^2}{x^2 +y^2}e^{-(x^2 +y^2)} \,dx\,dy$$

And converting this into polars gives:

$$\int_0^\infty r^2 e^{-r^2}\,dr \int_0^\infty\sin(\theta)\cos^2(\theta)\,d\theta $$

The integral w.r.t $r$ resolves to $\sqrt\pi/4$, however I am left with trying to solve the integral w.r.t $\theta$, which if solving it with substituting $u= \cos(\theta)$, requires me to evaluate $\cos(\infty)$, which does not converge.

Any help would be appreciated in either resolving this infinity or by a different approach which doesn't lead to this problem. Many thanks in advance.

Answer 1

When you convert the integral $\displaystyle \int_0^{\infty} \int_0^{\infty}\frac{yx^2}{x^2 +y^2}e^{x^2 +y^2}\ dx\ dy$ to polar form, you get $\displaystyle \int_0^{\frac{\pi}{2}} \int_0^{\infty} r^2e^{r^2}\sin\theta\cos^2\theta\ dr\ d\theta$. This happens because the rectangular integral is using the first quadrant for the limits of integration. In polar coordinates, the first quadrant only needs $0$ to $\dfrac{\pi}{2}$ for the angles.

Answer 2

The angle $\theta$ should be from $0$ to $\frac{\pi}{2}$ and so
$$\int_0^\infty r^2 e^{-r^2}dr \int_0^{\frac{\pi}{2}}\sin(\theta)\cos^2(\theta)d\theta$$ The first one is is given by Gamma function(Let $u=r^2$) and the second is direct (Let $u=\cos \theta$)