TITLE: How prove this limit $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{i+j}{i^2+j^2}=\frac{\pi}{2}+\ln{2}$
QUESTION [12 upvotes]: show that: this limit
$$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\dfrac{i+j}{i^2+j^2}=\dfrac{\pi}{2}+\ln{2}$$
My try: 
$$I=\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}\dfrac{\dfrac{i}{n}+\dfrac{j}{n}}{\left(\dfrac{i}{n}\right)^2+\left(\dfrac{j}{n}\right)^2}=\int_{0}^{1}\int_{0}^{1}\dfrac{x+y}{x^2+y^2}dxdy?$$
This idea is true? and have other methods?

REPLY [6 votes]: Evaluation of the integral(after Thomas Andrews's calculation)
$$
\int_0^1 \log(1+y^2) dy= y\log(1+y^2)|_0^1 -\int_0^1 \frac{2y^2}{1+y^2} dy$$
$$=\log 2-\int_0^1 \frac{2y^2+2-2}{1+y^2} dy$$
$$=\log 2 - 2 + \frac{\pi}{2} $$
$$-\int_0^1 \log (y^2) dy = -\int_0^1 2\log y dy = 2\int_0^{\infty} e^{-x} dx = 2$$
Hence, the integral all together is 
$$\log 2 + \frac{\pi}{2}$$
Application of Stolz-Cesaro: 
If we can prove that the limit exists
$$
\lim_{n\rightarrow\infty}\left( \sum_{i=1}^n \frac{i+n}{i^2+n^2} +\sum_{j=1}^{n-1} \frac{n+j}{n^2+j^2}\right)  $$
then the answer is the same as the limit. 
Evaluation of this limit requires integral, but this time not a 'double integral'.
$$\lim_{n\rightarrow\infty}  \sum_{i=1}^n \frac{i+n}{i^2+n^2}$$
$$=\lim_{n\rightarrow\infty} \sum_{i=1}^n \frac{ \frac{i}{n}+1}{\frac{i^2}{n^2}+1} \frac{1}{n}$$
$$=\int_0^1 \frac{x+1}{x^2+1} dx$$
$$=\frac{\pi}{4}+\frac{1}{2}\log 2.$$
The other part can be dealt similarly, it has the same value as above. Thus, all together the answer becomes 
$$\frac{\pi}{2}+\log 2.$$