Study the function $$f_\alpha(x,y)=\frac{x^\alpha}{x^2+y^2},$$ with $\alpha \geq 0$ and the domain $D\subset \mathbb{R}^2$ that in polar coordinates is given by $0\leq r \leq 1$ and $0 \leq \theta \leq \frac{\pi}{4}.$
Determine the values of $\alpha$ for which the improper integral exists.
Update I've expressed the integral into polar coordinates which yields the following $$\iint_D \frac{(r\cos(\theta))^\alpha}{r^2}r\ drd\theta$$ which can be rewritten into $$\iint_D r^{\alpha-1}\cos(\theta)^\alpha\ drd\theta$$ But I don't know how to proceed.
HINT 1 In polar coordinates you have $f_\alpha(r,\theta)=r^{\alpha-2}\cos^{\alpha}\theta$ so you have to discuss the to integrals $$ \iint_D f_\alpha(r,\theta)r\mathrm d r\mathrm d\theta=\int_0^1 r^{\alpha-1}\mathrm d r\int_0^{\pi/4} \cos^{\alpha}(\theta)\,\mathrm d \theta $$
HINT 2 $$ \int_0^1 \frac{1}{x^p}\mathrm d x=\begin{cases} +\infty & \text{for } p\leq 1\\ \frac{1}{1-p} & \text{for } p> 1 \end{cases} $$
HINT 3 For the integral with $\cos\theta$, use the fact that for $\theta\in[0,\pi/4]$ we have $\frac{1}{\sqrt{2}}\le\cos\theta\le 1$.