Deriving a relation between the gamma function and the Riemann zeta function.

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$\newcommand{\dd}{\mathrm{d}}$

The Mellin transform of a function $f$ is given by

\[\begin{aligned} \left\{\mathscr{M}f\right\}(s)=\phi(s)=\int_0^{\infty}x^{s-1}f(x)\dd x \end{aligned}\]

For the function

\[\begin{aligned} f(x)=\frac{1}{e^x - 1} \end{aligned}\]

we have

\[\begin{aligned} \phi(s)&=\int_0^{\infty}\frac{x^{s-1}}{e^x - 1}\dd x \\ &=\sum_{n \geq1}\int_{0}^{\infty}x^{s-1}e^{-nx}\dd x \end{aligned}\]

Making the substitution $u=nx$ gives

\[\begin{aligned} \phi(s)&=\sum_{n \geq1}\frac{1}{n^s}\int_{0}^{\infty}u^{s-1}e^{-u}\dd u \\ &=\zeta(s)\Gamma(s) \end{aligned}\]

Equating this to the above integral gives the desired result:

\[\begin{aligned} \zeta(s)\Gamma(s)&=\int_{0}^{\infty}\frac{x^{s-1}}{e^x - 1}\dd x \\ \zeta(s)&=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{e^x - 1}\dd x \end{aligned}\]

Another method of deriving this result is by starting with the Gamma function itself:

\[\begin{equation*} \begin{aligned} \Gamma(s) = \int_{0}^{\infty} t^{s-1} e^{-t} \dd t \end{aligned} \end{equation*}\]

Making the substitution $t = nu$ with $\dd t = n\dd u$, we have

\[\begin{equation*} \begin{aligned} \Gamma(s) &= \int_{0}^{\infty} (nu)^{s-1} e^{-nu} \dd u \\ &= \int_{0}^{\infty} n^s u^{s-1} e^{-nu} \dd u \end{aligned} \end{equation*}\]

The $n^s$ term can be brought out of the integral, and multiplying both sides by $\frac{1}{n^s}$ gives,

\[\begin{equation*} \begin{aligned} \Gamma(s) \frac{1}{n^s} = \int_{0}^{\infty} u^{s-1} e^{-nu} \dd u \end{aligned} \end{equation*}\]

Taking the sum over both sides gives us the Riemann zeta function on the left,

\[\begin{equation*} \begin{aligned} \Gamma(s) \sum_{n=1}^{\infty} \frac{1}{n^s} = \sum_{n=1}^{\infty} \int_{0}^{\infty} u^{s-1} e^{-nu} \dd u \\ \Gamma(s) \zeta(s) = \int_{0}^{\infty} u^{s-1} \sum_{n=1}^{\infty} e^{-nu} \dd u \\ \end{aligned} \end{equation*}\]

Seeing that the $e^{-nu}$ term is just $({e^{-u}})^n$, we can rewrite the infinite geometric series:

\[\begin{equation*} \begin{aligned} \Gamma(s)\zeta(s) = \int_{0}^{\infty} u^{s-1} \left( \frac{1}{1 - e^{-u}} - 1 \right) \dd u \end{aligned} \end{equation*}\]

Note that $1$ must be subtracted as the summation starts from $n=1$ rather than $n=0$. With some rearranging and manipulation we get,

\[\begin{equation*} \begin{aligned} \Gamma(s) \zeta(s) &= \int_{0}^{\infty} u^{s-1} \left( \frac{1}{1 - e^{-u}} - \frac{1 - e^{-u}}{1 - e^{-u}} \right) \dd u \\ &= \int_{0}^{\infty} u^{s-1} \left( \frac{e^{-u}}{1 - e^{-u}} \right) \dd u \\ &= \int_{0}^{\infty} u^{s-1} \left( \frac{1}{e^u - 1} \right) \dd u \end{aligned} \end{equation*}\]

And so, we have the desired result:

\[\begin{equation*} \begin{aligned} \zeta(s) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}}{e^u - 1} \dd u \end{aligned} \end{equation*}\]