It is well known, that the Zeta function has this integral formula for $\Re z>1$:
$\Gamma(z)\zeta(z)=\int _0^{\infty}\frac{t^{z-1}}{e^t-1} \ dt$
In the proof (attached in the link), we can interchange the integration and the sum: $\sum_{n=1}^{\infty}n^{-z}\Gamma(z)=\sum_{n=1}^{\infty}\big(\int _0^{\infty}t^{z-1}e^{-nt} \ dt \big)=\int _0^{\infty}\big(\sum_{n=1}^{\infty} t^{z-1}e^{-nt} \big) \ dt$. Can anybody tell me why we can interchange the integration and the sum? How can we prove this? Which theorem justifies this? (Maybe Lebesgue's dominated convergence theorem?)
For $\Re(s) > 1$ $$\sum_{n=1}^\infty n^{-s}\Gamma(s) = \lim_{N\to \infty}\sum_{n=1}^N \int_0^\infty x^{s-1}e^{-nx}dx= \lim_{N\to \infty}\int_0^\infty x^{s-1}\frac{1-e^{-Nx}}{e^x-1}dx$$
$$= \lim_{N\to \infty}\int_0^\infty \frac{x^{s-1}}{e^x-1}dx+O(e^{-N^{1/2}}\int_{N^{-1/2}}^\infty \frac{|x^{s-1}|}{e^x-1}dx)+O(\int_0^{N^{-1/2}} \frac{|x^{s-1}|}{e^x-1}dx)$$ $$ = \int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$