Let $$ \xi\left(z\right) = z\left(z - 1\right) \pi^{-z/2}\,\,\Gamma\left(\frac{z}{2}\right)\zeta\left(z\right), $$ where $\zeta(z)$ is the Riemann zeta function.
After some manipulations we may write: $$\xi(z)=\int_{-\infty}^{+\infty}\Phi(u)\exp(izu)du$$ Where $\Phi(u)=2e^{2u}\{3\psi'(e^{2u})+2e^{2u}\psi''(e^{2u})\}e^{\frac{u}{2}}$
and $\psi(x)=\sum_{n=1}^{\infty}\exp(-\pi n^2x)$
And this is well known result.
Here is what i have done. Let $F(u)=\psi(e^{2u})$, then after few calculations we get: $\Phi(u)=(F'(u)+F''(u))e^{\frac{u}{2}}$.
Lets consider a Fourier transform of $F$. This is: $$W(z):=\int_{-\infty}^{+\infty}F(u)\exp(izu)du$$
Again after some calculation and using properties of Fourier transform, we get: $$\xi(z-\frac{1}{2i}) = \int_{-\infty}^{+\infty}(F'(u)+F''(u))e^{\frac{u}{2}}\exp(i(z-\frac{1}{2i})u)du=$$ $$=\int_{-\infty}^{+\infty}(F'(u)+F''(u))\exp(izu)=izW(z)+(iz)^2W(z)$$
This implies that $z= +\frac{1}{2i}$ and $z=i+\frac{1}{2i}$ are zeros of function $\xi(z)$
The question:
Where is a mistake?
Thank you in advance
Regards.
Let $z=w-i/2$ then by your calculation, $$\xi(w)=i(w-i/2)W(w-i/2)-(w-i/2)^2W(w-i/2),$$ and as you say it follows that $$\xi(-i/2)=i(-i)W(-i)-(-i)^2W(-i)=W(-i)-W(-i)=0$$ and $$\xi(i-i/2)=i(i-i)W(i-i)-(i-i)^2W(i-i)=0,$$ so algebraically $-i/2$ and $i-i/2$ do appear to be roots of $\xi$.
Maybe the problem lies in the convergence of the integral $W(z)$. For example, $$W(-i)=\int_{-\infty}^\infty\sum_{n=1}^\infty\exp(-\pi n^2 e^{2u})\exp(u)du=\int_{-\infty}^\infty\frac{1}{2} e^u \left(\vartheta _3\left(0,e^{-e^{2 u} \pi }\right)-1\right)du,$$ where $\vartheta _3\left(0,e^{-e^{2 u} \pi }\right)$ is an Elliptic Theta function. Plotting this integrand (below) over $\mathbb{R}$ indicates that the integral may diverge. I guess you'd need to prove that $\frac{1}{2} e^u \left(\vartheta _3\left(0,e^{-e^{2 u} \pi }\right)-1\right)\to1/2$ as $u\to-\infty$.