For which values $\alpha \in \mathbb{R}$ is $f$ integrable?
$$f: \mathbb{R}^2 \rightarrow \mathbb{R} : f(x, y) = x \frac{\ln(1 + x^2 + y^2)}{(x^2 + y^2)^\alpha} $$
if $ (x,y) \neq (0, 0) $ and $ f(0, 0) = 2015$.
I tried to solve this question by switching variables to polar coordinates but I couldn't.
$f$ is integrable if and only if ${3\over 2}<\alpha < {5\over 2}$. Polar coordinates are actually useful. Write $(x,y)=r(\sin\theta,\cos\theta)$ and then you have $$f(x,y)dxdy=f(r,\theta)rdrd\theta=\sin\theta r^{2-2\alpha} \log(1+r^2)drd\theta$$ $f$ is clearly continuous away from the origin so we need only check what happen near zero and "near" infinity.
In a punctured neighborhood of the origin, the dependence of $f$ on $r$ is of the order of $r^{4-2\alpha}$ (because $\log(1+r^2)/r^2\to 1$ as $r \to 0$) so for $f$ to be integrable near the origin we need $4-2\alpha>-1$. Now, when $r\to\infty$ we need the power of $r$ to be smaller than $-1$, that is, $2-2\alpha < -1$. Combining these two inequalities for $\alpha$ gives the result as stated above.