Consider $F(z)$ a function such that $\overline{F(z)}=F(\overline{z})$, with no pole, decreasing faster than any power $\frac{1}{z}$ when z is imaginary going to $_{-}^{+}i \infty$. I define the following integral on the line $Re(z)=\frac{1}{2}$:
$$I = \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} F(z)F(1-z) (1-\sigma n^{-z})(1- \sigma n^{z-1}) dz$$
(with n an integer and $\sigma=_{-}^{+}1$)
By obvious change of variable we see that $I$ is real positive as we have :
$$I = i \int_{-\infty}^{+\infty} |F(\frac{1}{2}+it)|^2 |(1-\sigma n^{-\frac{1}{2}-it})|^2 dt$$
Now moving the integral to the line $\frac{3}{2}$ we have by therorem of residue (as integration on horizontal line between $\frac{1}{2} +it$ and $\frac{3}{2} +it$ will tend to zero when $t \to \infty$):
$$I = - i\int_{-\infty}^{+\infty} F(\frac{3}{2}+it)F(-\frac{1}{2}-it) (1-\sigma n^{-\frac{3}{2}-it})(1- \sigma n^{\frac{1}{2}+it}) dt$$
$$ = - i\int_{-\infty}^{+\infty} F(\frac{3}{2}+it)F(-\frac{1}{2}-it) (1+\sigma^2 n^{-1}- \sigma n^{-\frac{3}{2}-it} -\sigma n^{\frac{1}{2}+it}) dt$$
under this form for $n \to \infty$ it seems that the following term is bounded :
$$\int_{-\infty}^{+\infty} F(\frac{3}{2}+it)F(-\frac{1}{2}-it) (1+n^{-1}- \sigma n^{-\frac{3}{2}-it}) dt$$
Whereas the other one tends to + or - infinity for $n\to \infty$ depending on $\sigma$ :
$$\int_{-\infty}^{+\infty} F(\frac{3}{2}+it)F(-\frac{1}{2}-it) (\sigma n^{\frac{1}{2}+it}) dt$$
To estimate the limit of above integral we write $n^{\frac{1}{2}+it}= n^{\frac{1}{2}} e^{it \ln(p)}$ we see that the integral above is a Fourier transform taken at $ln(p)$, we then use that :
$$Lim_{x\to \infty}\int_{-\infty}^{+\infty} G(t) e^{itx} dt \sim K \frac{1}{x^2} $$
So I found an incoherence with an positive integral equal to a negative one ! I cannot find the mistake? Any help will be appreciated.
The integral $$ I = \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} F(z)F(1-z) (1-\sigma n^{-z})(1- \sigma n^{z-1}) dz $$ is not equal to $$\int_{-i\infty}^{+i\infty} |F(\frac{1}{2}+it)|^2 |(1-\sigma n^{-\frac{1}{2}-it})|^2 dt$$ but equal to $$ i\int_{-\infty}^{+\infty} |F(\frac{1}{2}+it)|^2 |(1-\sigma n^{-\frac{1}{2}-it})|^2 dt ,$$ since $dz=idt$ when $z=\frac{1}{2}+it.$
Addendum:
Now moving the integral to the line $\frac{3}{2}$ we have \begin{align} I &=i\int_{-\infty}^{+\infty} F(\frac{3}{2}+it)F(-\frac{1}{2}-it) (1-\sigma n^{-\frac{3}{2}-it})(1- \sigma n^{\frac{1}{2}+it}) dt\\ &=i\int_{-\infty}^{+\infty} F(\frac{3}{2}+it)F(-\frac{1}{2}-it) (1+\sigma^2n^{-1}- \sigma n^{-\frac{3}{2}-it} -\sigma n^{\frac{1}{2}+it}) dt. \end{align} since $$ \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}=\int_{\frac{3}{2}-i\infty}^{\frac{3}{2}+i\infty} $$ and $dz=idt$ when $z=\frac{3}{2}+it.$