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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?

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The obvious problem with the double integral is that it has problems near $(0,0)$, so that this is an improper integral. – Thomas Andrews Jan 23 '14 at 3:19
The function is $O(1/(x+y))$, so the singularity is under control for purposes of integration, a $1/r$ blowup on area $r^2$ for $r \to 0$. @ThomasAndrews – zyx Jan 23 '14 at 3:52
I think that the methoid you propose in you post is both simple and elegant. I tried a few thinks, in particular eveluating the most inner summation for a given value of $i$; this leads to awful results involving for PolyGamma functions. As a result, the outer summation cannot be expressed analytically. As a conclusion from my side, I enjoy your method and I congratulate you for its simplicity. – Claude Leibovici Jan 23 '14 at 4:32

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 $$\frac{\pi}{2}+\log 2.$$

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How was Stolz-Cesaro used to get from a two-variable sum to a pair of one-variable sums? – zyx Jan 23 '14 at 19:26
$\lim a_n/n = \lim (a_n-a_{n-1})$ if the limit on the right exists. Here $a_n=\sum_1^n\sum_1^n (i+j)/(i^2+j^2)$. – i707107 Jan 23 '14 at 21:22
Why is @zyx asking about Stolz-Cesaro? Someone asked about that in my answer, too, yet it is not mentioned in the question. – Thomas Andrews Jan 23 '14 at 23:22
In a comment it was stated that there was a hint to solve the problem with Stolz-Cesaro, this answer used it, and my question (now answered) was to which pair of sequences the theorem was applied. @ThomasAndrews – zyx Jan 23 '14 at 23:39
@zyx Where did the OP say that? Was the comment deleted? Why didn't anybody suggest that OP put that in the question? – Thomas Andrews Jan 23 '14 at 23:41

[Incomplete answer]

Probably a good place to start is to note: $$\int_0^1\int_0^1\frac{x}{x^2+y^2}\,dx\,dy = \int_0^1\int_0^1\frac{y}{x^2+y^2}\,dx\,dy$$

by symmetry, assuming the integral exists.

So you really need to compute $$\int_0^1\int_0^1 \frac{2x}{x^2+y^2}\,dx \,dy$$

That looks easier to compute.

$$\int_0^1\frac{2x}{x^2+y^2} dx = \log (1+y^2)-\log(y^2) = \log(1+y^{-2})$$

That leads to $$\int_0^1 \log(1+y^{-2})\,dy$$

Not sure if that's going to do any good.

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It is said can use Stolz–Ces`aro lemma,and then can use without Double integral – china math Jan 23 '14 at 3:49

(1) Integral is finite.

The function in the integral is $\leq \frac{2}{x+y}$. The integral of $1/(x+y)$ on the square is bounded, because $\frac{1}{x+y} \in [N,N+1)$ for points in the square with $x+y \in [1/(N+1), 1/N]$ is (for large $N$) a trapezoid with height approximately $1/N^2$ and width of approximately $1/N$, for a total contribution to the integral of about $N \times \frac{1}{N^3} (1 + o(1)) \sim \frac{1}{N^2} $. This converges when summed on integer $N \geq n_0$.

(2) Sum converges to the integral.

The sum is a "minimum" Riemann sum for the integral on a partition of the square into $n^2$ equal small. The "maximum" Riemann sum on the same squares except the one containing $(0,0)$ (whose contribution to the sum and to the integral both converge to $0$, and can be ignored) is the same sum for $i,j < n$, plus a small modification along the boundary of the square ($ij=0$ or $i=n$ or $j=n$). These changes to the sum are essentially line integrals of a the function along the sides of the square, weighted by $1/n$, and the modifications go to $0$.

(3) Evaluation of integral.

This can be done using polar coordinates, and maybe by easier methods. I did not work it out, since parts 1+2 already identified what the limit "is" and showed convergence.

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(Integral now done in one of the other answers) – zyx Jan 23 '14 at 23:40

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