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Aymen Hafeez

The Mellin transform of a function ff is given by

{Mf}(s)=ϕ(s)=∫0∞xs−1f(x)dx \begin{aligned} \left\{\mathscr{M}f\right\}(s)=\phi(s)=\int_0^{\infty}x^{s-1}f(x)\text{d} x \end{aligned}

For the function

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

we have

ϕ(s)=∫0∞xs−1ex−1dx=∑n≥1∫0∞xs−1e−nxdx \begin{aligned} \phi(s)&=\int_0^{\infty}\frac{x^{s-1}}{e^x - 1}\text{d} x \\ &=\sum_{n \geq1}\int_{0}^{\infty}x^{s-1}e^{-nx}\text{d} x \end{aligned}

Making the substitution u=nxu=nx gives

ϕ(s)=∑n≥11ns∫0∞us−1e−udu=ζ(s)Γ(s) \begin{aligned} \phi(s)&=\sum_{n \geq1}\frac{1}{n^s}\int_{0}^{\infty}u^{s-1}e^{-u}\text{d} u \\ &=\zeta(s)\Gamma(s) \end{aligned}

Equating this to the above integral gives the desired result:

ζ(s)Γ(s)=∫0∞xs−1ex−1dxζ(s)=1Γ(s)∫0∞xs−1ex−1dx \begin{aligned} \zeta(s)\Gamma(s)&=\int_{0}^{\infty}\frac{x^{s-1}}{e^x - 1}\text{d} x \\ \zeta(s)&=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{e^x - 1}\text{d} x \end{aligned}

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

Γ(s)=∫0∞ts−1e−tdt\begin{equation*} \begin{aligned} \Gamma(s) = \int_{0}^{\infty} t^{s-1} e^{-t} \text{d} t \end{aligned} \end{equation*}

Making the substitution t=nut = nu with dt=ndu\text{d} t = n\text{d} u, we have

Γ(s)=∫0∞(nu)s−1e−nudu=∫0∞nsus−1e−nudu\begin{equation*} \begin{aligned} \Gamma(s) &= \int_{0}^{\infty} (nu)^{s-1} e^{-nu} \text{d} u \\ &= \int_{0}^{\infty} n^s u^{s-1} e^{-nu} \text{d} u \end{aligned} \end{equation*}

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

Γ(s)1ns=∫0∞us−1e−nudu\begin{equation*} \begin{aligned} \Gamma(s) \frac{1}{n^s} = \int_{0}^{\infty} u^{s-1} e^{-nu} \text{d} u \end{aligned} \end{equation*}

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

Γ(s)∑n=1∞1ns=∑n=1∞∫0∞us−1e−nuduΓ(s)ζ(s)=∫0∞us−1∑n=1∞e−nudu\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} \text{d} u \\ \Gamma(s) \zeta(s) = \int_{0}^{\infty} u^{s-1} \sum_{n=1}^{\infty} e^{-nu} \text{d} u \\ \end{aligned} \end{equation*}

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

Γ(s)ζ(s)=∫0∞us−1(11−e−u−1)du\begin{equation*} \begin{aligned} \Gamma(s)\zeta(s) = \int_{0}^{\infty} u^{s-1} \left( \frac{1}{1 - e^{-u}} - 1 \right) \text{d} u \end{aligned} \end{equation*}

Note that 11 must be subtracted as the summation starts from n=1n=1 rather than n=0n=0. With some rearranging and manipulation we get,

Γ(s)ζ(s)=∫0∞us−1(11−e−u−1−e−u1−e−u)du=∫0∞us−1(e−u1−e−u)du=∫0∞us−1(1eu−1)du\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) \text{d} u \\ &= \int_{0}^{\infty} u^{s-1} \left( \frac{e^{-u}}{1 - e^{-u}} \right) \text{d} u \\ &= \int_{0}^{\infty} u^{s-1} \left( \frac{1}{e^u - 1} \right) \text{d} u \end{aligned} \end{equation*}

And so, we have the desired result:

ζ(s)=1Γ(s)∫0∞us−1eu−1du\begin{equation*} \begin{aligned} \zeta(s) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}}{e^u - 1} \text{d} u \end{aligned} \end{equation*}