I want to proof that I can switch this Sum and Integral
$\sum\limits_{n=1}^\infty\int\limits_{0}^\infty t^{z-1} e^{-nt}dt~~$ for $~ 1 < Re(z) $
to sum it after over n. I tried to use the geometric series by the estimate
$\vert{t^{z-1} e^{-nt}}\vert =\vert t^{x+iy-1} e^{-nt}\vert = \vert e^{ln(t)(x+iy-1)} e^{-nt}\vert =\vert e^{ln(t)(x-1)} e^{-nt}e^{iy\cdot ln(t)}\vert \leq \vert e^{ln(t)(x-1)} e^{-nt}\vert \leq \vert \frac{t^{x-1}}{e^{t}}\vert \leq t^{x-1}$
But this $t^{x-1}\leq 1$ is only true for $t\leq 1$. Where are my mistakes or are there any other ideas for the proof?
The integrand is not a positive function so there is some justification needed here. We can switch the integral and the sum if we know that $\sum_n \int |t^{z-1} e^{-nt}|dt<\infty$. This reduces to $\sum_n \int t^{x-1} e^{-nt}dt<\infty$. Here we can switch the sum and the integral by Tonelli's Theorem, so we now have to show that $\int t^{x-1} \frac {e^{-t}}{ 1-e^{-t}} dt <\infty$. I leave it to you to show that the integrand here is dominated by $Ce^{-t/2}$ for some constant $C$. This completes the proof.