I am trying to compute the following integral:
\begin{equation} \int_{x_1,x_2,x_3,y_1,y_2,y_3}\sum_{\vec{0}\neq\vec{n}\in Z^3}\sum_{\vec{0}\neq\vec{m}\in Z^3} \frac{m_1n_1}{\vec{n}^2\vec{m}^2}e^{i2\pi\vec{n}\cdot\left(x_1,x_2,y_3\right)+i2\pi\vec{m}\cdot\left(x_1,y_2,x_3\right)} dx^1dx^2dx^3dy^1dy^2dy^3\, , \end{equation}
where the integral goes from $0$ to $1$ for all variables $x^i$, $y^i$; $\vec n=(n_1,n_2,n_3)$ and the same for $\vec m$. I have been reading about interchanging the sum and the integral and I am not sure if I can do it in this case. If this was possible, then I do know how to proceed, so my question is if I can first do the integral and afterwards the sum, this is: \begin{equation} \sum_{\vec{0}\neq\vec{n}\in Z^3}\sum_{\vec{0}\neq\vec{m}\in Z^3}\frac{m_1n_1}{\vec{n}^2\vec{m}^2}\int_{x_1,x_2,x_3,y_1,y_2,y_3} e^{i2\pi\vec{n}\cdot\left(x_1,x_2,y_3\right)+i2\pi\vec{m}\cdot\left(x_1,y_2,x_3\right)} dx^1dx^2dx^3dy^1dy^2dy^3\, . \end{equation}
If the answer is no, what is happening then? Is it the result divergent -I know the series is-? Can I just argue that I am 'renormalising' the result in some sense?