Computing $\sum{\frac{1}{m^2+n^2}}$ I want to prove that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=2\pi\log R+O(1)$ as $R\rightarrow\infty$.
For this, I'm trying to approximate the sum by using the integral $\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy=\int_{0}^{2\pi}\int_{1}^{R}{\frac{1}{r}}drd\theta=2\pi\log R$
How can I show that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy$ as $R\rightarrow\infty$ rigorously?
Any help will be appreciated.

Just an idea:
Let's use
$$x-1<[x]\leq x$$
$$\sum \frac{1}{m^2+n^2}=\int \frac{1}{[x]^2+[y]^2}$$
Then
$$\int \frac{1}{x^2+y^2}\leq\int \frac{1}{[x]^2+[y]^2}<\int \frac{1}{(x-1)^2+(y-1)^2}$$
And hopefully we can show
$$\lim_{R\to\infty}\int \frac{1}{x^2+y^2}=\lim_{R\to\infty}\int \frac{1}{(x-1)^2+(y-1)^2}$$
?
