Integral of a special Fourier Serie into the hipercube of s dimension. Given the numbers, $u\in\mathbb{R}^s$, $\alpha>1$ and $s>1$. 
If we have the below Fourier Serie:
$f_{\alpha}(u)=\sum\limits_{h\in\mathbb{Z}^s}\frac{1}{r(h)^{\alpha}}\exp^{2 \pi i \left<h,u\right>}$, 
where, $r(h)=\prod\limits_{i=1}^{s}max(1,\vert h_{i} \vert )$. 
Then, the integral of this function over the inside of the hipercube of s dimension has the value of one. $\int_{[0,1)^{s}}f_{\alpha}(u) du=1$.
 I have serious doubts about it. Someone could help me?.
Thanks.
Reference: Pricing Options Using Lattice Rules. Phelim P. Boyle, Yongzeng Lai, and Ken Seng Tan. 
 A: Since we can write $f_{\alpha}(u)=\prod\limits_{i=1}^sF_{\alpha}(u_i)$ where $F_{\alpha}(x)=\sum\limits_{n\in\mathbb{Z}}\frac{\exp(2\pi inx)}{\max(1,|n|)^{\alpha}}$, it is enough to check that $\int_0^1F_{\alpha}(x)dx=1$.
\begin{equation}\int_0^1F_{\alpha}(x)dx=\underbrace{\sum\limits_{n\in\mathbb{Z}\atop n\ne0}\frac{1}{\max(1,|n|)^{\alpha}}\int_0^1e^{2\pi inx}dx}_{A} + \int_0^11dx=A+1.\end{equation}
Now we see that $A=0$ by pairing the terms $n$ and $-n$,
\begin{align}A=&\sum\limits_{n\in\mathbb{Z}\atop n\ne0}\frac{1}{\max(1,|n|)^{\alpha}}\int_0^1e^{2\pi inx}dx=\sum\limits_{n\in\mathbb{N}}\frac{1}{|n|^{\alpha}}\int_0^1\left(e^{2\pi inx}+e^{-2\pi inx}\right)dx\\
&= \sum\limits_{n\in\mathbb{N}}\frac{2}{|n|^{\alpha}}\int_0^1\cos(2\pi n x)dx=0.\end{align}
A: Thank you very much for your right answer! 
At the beginning I did not realize that $f_{\alpha}(u)$ can be put as a product of $F_{u_i}$ with $u_i\in \mathbb{Z}$, so I had problems to get the solution. In that way, you have reduced my problem to one dimension easier.
But you still have not explained the detail at the begining of your answer to complete the solution.  
I think that is like this,
\begin{equation}
f_\alpha(u)=\sum_{h\in\mathbb{Z}^s}\frac{1}{h(u)^\alpha} e^{2\pi i <h,u>}=
\sum_{h\in\mathbb{Z}^s} \Bigg( \bigg( \prod_{i=1}^{s}\max{(1,|h_i|)} \bigg)^{-\alpha} \exp{\bigg(2\pi i (h_1u_1+\ldots+h_su_s)\bigg)} \Bigg)=
\end{equation}
\begin{equation}
=\sum_{h\in\mathbb{Z}^s} \Bigg( \frac{ \exp{(2\pi i h_1u_1)}\cdots \exp{(2\pi i h_s u_s)} }{ \max{(1,|h_1|)}^{\alpha}\cdots\max{(1,|h_s|)}^{\alpha}   } \Bigg)=
\end{equation}
\begin{equation}
=\sum_{h_1 \in\mathbb{Z}} \bigg(\frac{exp(2\pi i h_1u_1)}{\max{(1,|h_1|)}^{\alpha}} \bigg) \cdots \sum_{h_s\in\mathbb{Z}}\bigg(\frac{exp(2\pi i h_su_s)}{\max{(1,|h_s|)}^{\alpha}} \bigg)= \prod_{i=1}^{s} F_\alpha (u_i)
\end{equation}
Calling $F_\alpha,$
\begin{equation}
F_\alpha (x):=\sum_{h=-\infty}^{\infty}\frac{exp(2\pi i h x)}{\max{(1,|h|)}^{\alpha}}
\end{equation}
Therefore, it becomes one dimension problem. 
Thanks a lot.
