Let $ D=\{(x,y) \in \mathbb{R}^2 : x>0, y>0\}$. Show that the function $$f(x,y)={1 \over x^2} \sum_{n=1}^{\infty}{\int_x^y{\sqrt{t} \over {1+ ({t \over x} -n)^2}}} dt$$ is well defined on $D$. Determinate if $f$ is differentiable on $D$ and if $f \in L^1(D)$
My work:
Observation: $s=t/x$
$f(x,y)={1 \over \sqrt{x}} \sum_{n=1}^{\infty}{\int_1^{y/x}{\sqrt{s} \over {1+ ({s} -n)^2}}} ds={1 \over \sqrt{x}} {\int_1^{y/x} \sum_{n=1}^{\infty}{\sqrt{s} \over {1+ ({s} -n)^2}}} ds$
$f$ is well defined on $D$ : for $y>x>0$
$0 \le f(x,y) \le {\sqrt{y} (y-x) \over x^2} \sum_{n=1}^\infty{1 \over {1+(1-n)^2}} < +\infty$;
Similarly for $x>y>0$ we have $-\infty \le f(x,y) \le 0$;
For $x=y $ we have $f(x,y)=0$.
$F \notin L^1(D) $ : $x= \rho\cos{\theta}, y=\rho \sin{\theta}$ $\int_{0<x<y}|f|=\int_0^\infty{\int_{\pi/4}^{\pi/2}{1 \over {\sqrt{\rho\cos{\theta}}}}\int_1^{\tan{\theta}}\sum_{n=1}^\infty{\sqrt{s} \over {1+(s-n)^2}}ds\rho d\theta d\rho}$ and for the Tonelli' s theorem $f$ doesn't converge because $\int_0^\infty\sqrt{\rho} d\rho=+\infty$
I don't know how to establish the differentiability.