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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?)

The proof from Garling book

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For $\Re(s) > 1$ $$\sum_{n=1}^\infty n^{-s}\Gamma(s) = \lim_{N\to \infty}\sum_{n=1}^N \int_0^\infty x^{s-1}e^{-nx}dx= \lim_{N\to \infty}\int_0^\infty x^{s-1}\frac{1-e^{-Nx}}{e^x-1}dx$$

$$= \lim_{N\to \infty}\int_0^\infty \frac{x^{s-1}}{e^x-1}dx+O(e^{-N^{1/2}}\int_{N^{-1/2}}^\infty \frac{|x^{s-1}|}{e^x-1}dx)+O(\int_0^{N^{-1/2}} \frac{|x^{s-1}|}{e^x-1}dx)$$ $$ = \int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$

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