In the chatroom we discussed about the asymptotic of $\displaystyle \sum\_{i = 1}^n \sum\_{j = 1}^n \frac1{i^2+j^2}$, and if we think of the inverse tangent integral, it's easy to see that $\displaystyle \sum\_{i = 1}^n \sum\_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}\_2(n)\approx \frac{\pi}{2} \log(n)$ where I used  

the well-known relation $\displaystyle \operatorname{Ti}\_2(x)-\operatorname{Ti}\_2(1/x)=\frac{\pi}{2}\operatorname{sgn}(x) \log|x|$. At this point, @robjohn posed the following limit

$$\lim\_{n\to\infty} \left(\displaystyle \sum\_{i = 1}^n \sum\_{j = 1}^n \frac1{i^2+j^2}-\frac{\pi}{2} \log(n)\right)$$
that looks like a pretty tough limit. Using $\coth(z)$ one can see the limit is approximately $-\dfrac{\pi^2}{12}$,  

but how about finding a way to precisely compute the limit?