Compute a multiple integral problem. My coursebook on integral calculus has a question: 


*

*By changing into polar coordinates, show that:
$$ \int_0^a\int_y^a \frac{1}{x^2+y^2}  dx dy = \frac{\pi a}{4} $$


This doesn't look good to me. When we take $y$ approaching $0$ (our lower bound on $y$), since value of $y$ is a lower limit for $x$, $x$ too approaches $0$ on the lower limit. In other words, our integral should account for the region upto and including (0,0). But, in such case, the value of function on integral grows with no bound (tends to $\infty$). Hence, the finite value the question asks for can't be shown.  
What am I missing here?  
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int_{0}^{a}\int_{y}^{a}{1 \over x^{2} + y^{2}}\,\dd x\,\dd y =
{\pi a \over 4}:\ {\large ?}}$.

\begin{align}
\int_{0}^{a}\int_{y}^{a}{1 \over x^{2} + y^{2}}\,\dd x\,\dd y & =
\int_{0}^{a}\int_{0}^{a}{\bracks{x > y} \over x^{2} + y^{2}}\,\dd x\,\dd y
\\[5mm] & =
{1 \over 2}\braces{%
\int_{0}^{a}\int_{0}^{a}{\bracks{x > y} \over x^{2} + y^{2}}\,\dd x\,\dd y +
\int_{0}^{a}\int_{0}^{a}{\bracks{y > x} \over y^{2} + x^{2}}\,\dd y\,\dd x}
\\[5mm] & =
{1 \over 2}\int_{0}^{a}\int_{0}^{a}{\dd x\,\dd y \over x^{2} + y^{2}} =
{1 \over 2}\int_{0}^{a}{1 \over y}\int_{0}^{a/y}{\dd x \over x^{2} + 1}\,\dd y
\\[5mm] & =
{1 \over 2}\int_{0}^{a}{1 \over y}\,\arctan\pars{a \over y}\,\dd y\qquad
\pars{~\mbox{It}\ \underline{diverges}\ 'logaritmically'~}
\end{align}
