Let $s=\sigma+it$ is the complex variable, in the Laplace transform of the Riemann zeta-function we consider the following integral
$ \int\limits_{2-i\infty}^{2+i\infty}\Gamma^k(z)\zeta^{2k}(z)\left(\ (-1)^{-m}(2\pi i)^k e^{-is}\right)^{-z}dz $
$k$ $\in$ $\mathbb{N}$ and
$m=0,1,...,\frac {k-1}{2}$
To figure out the region of absolutely convergence of this integral I changed the variable
$z=2+i\omega$ an get two integrals
$ \int\limits_{-\infty}^{\infty}\Gamma^k(2+i \omega)\zeta^{2k}(2+i \omega)\left(\ (-1)^{-m}(2\pi i)^k e^{-is}\right)^{-i \omega}d\omega=\\ \int\limits_{0}^{\infty}\Gamma^k(2+i \omega)\zeta^{2k}(2+i \omega)\left(\ (-1)^{-m}(2\pi i)^k e^{-is}\right)^{-i \omega}d\omega+\int\limits_{0}^{\infty}\Gamma^k(2-i \omega)\zeta^{2k}(2-i \omega)\left(\ (-1)^{-m}(2\pi i)^k e^{-is}\right)^{i \omega}d\omega\\ =I_1+I_2 $
For the first integral we have estimates
$ \Gamma^k(2\pm i \omega)\sim e^{-\frac{\pi}{2}|\omega|k} $
$ |(2\pi i)^{-ki\omega}|=|e^{-ki\omega\log(2\pi i)}|=e^{\omega\frac{\pi}{2}k} $
$ |(-1)^{m i\omega}|=e^{-m \pi \omega} $
and the region of absolute convergence of the first integral we get
$-\frac{\pi}{2}k-\sigma+\frac{\pi}{2}k-m\pi<0$
$\sigma>-m\pi$
But it isn' correct because we get the region of absolute convergence of the first integral depends on m and increasing. In fact if $m$- is odd $(-1)^{-m}=-1$ and if $m$- is even $(-1)^{-m}=1$, so the region of absolute convergence couldn't expand if we changing $m$.
Question: Where is the problem?
Moreover we have
$(-1)^{-m}=(-1)^{m}$
therefore in this case we get region $\sigma>m\pi$
I understand that this related with the path of logarithm but I don't know how to explain this mathematically?
For the second one we have estimates
$ |(2\pi i)^{ki\omega}|=|e^{ki\omega\log(2\pi i)}|=e^{-\omega\frac{\pi}{2}k} $
$ |(-1)^{-m i\omega}|=e^{m \pi \omega} $
and region of absolute convergence is
$-\frac{\pi}{2}k+\sigma-\frac{\pi}{2}k+m\pi<0$
$\sigma<(k-m)\pi$
I need to show that region of absolutely convergence of both integrals is
$\sigma \in (0,\frac{(k+1)\pi}{2})$
It is true?
For $k\ge 1$
$$\int\limits_{2-i\infty}^{2+i\infty}\Gamma^k(z)\zeta^n(z) e^{uz}dz$$ converges absolutely iff $|\Im(u)|< k \frac{\pi}2$.
$|\zeta^n(z)|$ is bounded above and below so we don't care of it. By Stirling on $\Re(z)\ge 2$ we have $\Gamma^k(z) = e^{k z \log z- z-\frac12 \log z+O(1)}$ and $\log(2+it) = \log(it)+O(1/t) = \log(|t|)+isign(t)\frac{\pi}2+O(1/t)$ so $ \Gamma^k(2+it) =e^{k \frac32 \log|t|-k\frac{\pi}2 |t| + O(1)}$