Approximating $\iint \frac{(x+y)^2}{(1-\exp(-x-y))(x^2+y^2)^2} \,dy\,dx$ Fix two positive numbers $x_{0}, y_{0} > 0$, and then also fix two very, very large numbers $x_{B}, y_{B}$. Consider the integral:
$$
I = \int_{x_{0}}^{x_{B}} \int_{y_{0}}^{y_{B}} \frac{(x+y)^2}{\big[ 1-\exp(-x-y) \big] (x^2+y^2)^2} \,dy\,dx
$$
I am interested in how this integral looks like as a function of $x_{B}, y_{B}$ since these are big numbers ''close to $\infty$''.
This integral diverges as $x_{B}, y_{B} \to \infty$ (also at the origin, but I am not interested in what is going on there). I am interested how this integral looks like for large $x_{B}, y_{B}$.
My approach: Since I am interested in the limit that both $x$ and $y$ are large, I think I can make the approximation that:
$$
I \approx \int_{x_{0}}^{x_{B}} \int_{y_{0}}^{y_{B}} \frac{(x+y)^2}{ (x^2+y^2)^2} \,dy\,dx
$$
This is because the exponential term dies away very quickly in the original integrand. Potential Issue: I think that in order to make use of the above approximation I have to also assume that not only $x_{B}, y_{B}$ are large, but also the lower limits $x_{0}, y_{0}$ are large.
I have found that:
$$
\frac{\partial^{2}}{\partial x\ \partial y} \bigg[ \frac{i}{2} \mathrm{Li}_{2}\left(i\frac{y}{x}\right) - \frac{i}{2} \mathrm{Li}_{2}\left(- i\frac{y}{x}\right) + \frac{1}{2} \ln( x^2 + y^2 ) \bigg] = \frac{(x+y)^2}{ (x^2+y^2)^2}
$$
Where $\mathrm{Li}_{2}$ is the dilogarithm. The derivative $\frac{\partial^{2}}{\partial y\ \partial x}$ is the same.
So my question is; can I make the following approximation?
$$
I \approx \frac{i}{2} \mathrm{Li}_{2}\left(i\frac{y_{B}}{x_{B}}\right) - \frac{i}{2} \mathrm{Li}_{2}\left(- i\frac{y_{B}}{x_{B}}) + \frac{1}{2} \ln( x_{B}^2 + y_{B}^2 \right)
$$
(Is it a good approximation in the sense that if I make $x_{B}, y_{B}$ large enough then any involvement of $x_{0}, y_{0}$ doesn't matter; I'm trying to look at the asymptotics in terms of  the upper limits$x_{B}, y_{B}$)
I am worried that this is not valid because of any mixing of the above antiderivative with the lower limits $x_{0}, y_{0}$
Furthermore, I haven't been able to check my approximation numerically because the integral diverges so slowly as $x_{B}, y_{B} \to \infty$ that I can't even do anything with Mathematica (it looks to be diverging logarithmically).
 A: So I had another idea, which is to convert to polar coordinates $x = R\cos(\theta)$ and $y = R \sin (\theta)$. Saying the the upper limit is approximately $\sqrt{x_{B}^2 + y_{B}^2}$ and the bottom limit is approximately $\sqrt{x_{0}^2 + y_{0}^2}$, we have roughly:
$$
I \ \approx \ \int\limits_{0}^{\pi/2} d \theta \int\limits^{\sqrt{x_{B}^2 + y_{B}^2}}_\sqrt{x_{0}^2 + y_{0}^2} R \cdot \frac{ \left( R \cos(\theta) + R \sin(\theta) \right)^2}{\left(1 - e^{- R \cos(\theta) - R \sin(\theta)}\right)( R^{2})^{2}} \ = \ 2 \int\limits_{0}^{\pi/2} d \theta \int\limits^{\sqrt{x_{B}^2 + y_{B}^2}}_\sqrt{x_{0}^2 + y_{0}^2} \frac{ \sin^2(\theta+\frac{\pi}{4})}{\left(1 - e^{- \sqrt{2} R \sin(\theta +\frac{\pi}{4})}\right)R} 
$$
But since we only care about very large $R$ (aka. making $x_{B}$, $y_{B}$, $x_{0}$, $y_{0}$ very large), the integral approximately looks like:
$$
I \ \approx \ 2 \int\limits_{0}^{\pi/2} d \theta \int\limits^{\sqrt{x_{B}^2 + y_{B}^2}}_\sqrt{x_{0}^2 + y_{0}^2} \frac{ \sin^2(\theta+\frac{\pi}{4})}{R} \ = \ 2\left(1+\frac{\pi}{2}\right) \bigg[ \ln \left( \sqrt{x_{B}^2 + y_{B}^2} \right) - \ln \left( \sqrt{x_{0}^2 + y_{0}^2} \right) \bigg]
$$
And saying that $x_{B}, y_{B} \gg x_{0}, y_{0}$ the above is roughly:
$$
I \ \approx \ \left(1+\frac{\pi}{2}\right) \ln \left( x_{B}^2 + y_{B}^2 \right)
$$
even though my integral now integrates over an annular region rather than the original box, this seems okay to me.
