Let $m$ and $j$ be two positive integers. Then do we have :
$$ \displaystyle \int_{m}^{m+1} \int_{j}^{j+1} \frac{1}{\lfloor x \rfloor ^2+ \lfloor y \rfloor ^2}\, \mathrm{d}x\,\mathrm{d}y = \frac{1}{x^2+y^2}$$
If this is true, how can I prove it ?
According to me, this is false because by Fubini we have : $$ \displaystyle \int_{m}^{m+1} \int_{j}^{j+1} \frac{1}{\lfloor x \rfloor ^2+ \lfloor y \rfloor ^2} \mathrm{d}x\mathrm{d}y = \int_{m}^{m+1} \frac{1}{y} \cdot \left(\tan^{-1}\left(\frac{j+1}{\sqrt{y}}\right)-\tan^{-1}\left(\frac{j}{\sqrt{y}}\right)\right) \,\mathrm{d}y$$
But I am not sure...
I'm not sure that I follow any of your attempts to integrate this formula, and I don't understand why you think it should be that $$\int_{m}^{m+1} \int_{j}^{j+1} \frac{1}{\lfloor x \rfloor^2 + \lfloor y \rfloor^2} \,\mathrm{d}x\,\mathrm{d}y = \frac{1}{x^2 + y^2}.$$ On the entire region of integration, the function that you are integrating is constant. This makes the integral relatively simple to evaluate. There is really no need to invoke Fubini.
On the interval $[j,j+1)$, the value of $\lfloor x \rfloor$ is constant and equal to $j$. On the interval $[m,m+1)$, the value of $\lfloor y \rfloor$ is constant and equal to $m$. Therefore \begin{align} \int_{m}^{m+1} \int_{j}^{j+1} \frac{1}{\lfloor x \rfloor^2 + \lfloor y \rfloor^2} \,\mathrm{d}x\,\mathrm{d}y &= \int_{m}^{m+1} \int_{j}^{j+1} \frac{1}{j^2 + m^2} \,\mathrm{d}x\,\mathrm{d}y \\ &= \frac{1}{j^2 + m^2} \int_{m}^{m+1} \int_{j}^{j+1} \,\mathrm{d}x\,\mathrm{d}y \\ &= \frac{1}{j^2 + m^2}, \end{align} which is constant.