Convergence integral Riemann zeta

Question

In the paper of Riemann and the book of Edwards I encountered the following representation for $\zeta$: $$ \int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx=\Pi(s-1)\zeta(s) $$ which due to the fast growth of $e^x$ converges at infinity, for $x$ tending to $0$ the integrand grows like $x^{s-2}$, and therefore due to this pole the integral converges for $Re(s)>1$. Then he considers the following integral,taken over the Hankel contour from infinity to along the real axis anti-clockwise around the pole and back to infinity. \begin{align*} & \int_\infty^\infty\frac{(-x)^{s-1}}{e^x-1} \, dx \\[8pt] = {} & (e^{i\pi s}-e^{-i\pi s}) \int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx \\[8pt] = {} & 2i\sin(\pi s) \Pi(s-1)\zeta(s) \end{align*} It is claimed that the left integral converges for all $s.$ However, I was unable to see this convergence. I suspect by going around $z=0$ we omitted the pole, and therefore we only have to look at the convergence at infinity. But around $z=0$, a limiting process still takes place.
How can we rigorously prove the convergence of the left integral?

Answer 1

I wasn't clear exactly where you were stuck in the proof, so I posted a (more-or-less) full proof.

We assume one easily proved lemma:

lemma 1: For $|z| < 1/2$, $\left|e^z -1\right| \geq |z|/2$.

Next, we define a version of the Hankel contour. For small positive $\delta$ select two points in the complex plane: \begin{align*} p_1&=\delta e^{i\delta} = \delta_x + i\delta_y\quad\text{where}\quad \delta_x = \delta\cos(\delta), \delta_y = \delta\sin(\delta)\\ p_2&=\delta e^{- i\delta} = \overline{\delta e^{i\delta}} = \delta_x - i\delta_y \end{align*} Our contour $C$ consists of three piecewise smooth curves $C_1+C_2+C_3$ defined as follows:

$C_1$ follows a path parallel to (and just above) the real axis from $\infty$ to $p_1$.
$C_2$ follows a counterclockwise circular path from $p_1$ to $p_2$, tracing the boundary of a circle that is centered at 0 with radius $\delta$. (Making just barely less than a full circle).
$C_3$ follows a path parallel to (and just below) the real axis from $p_2$ to $\infty$.

Lemma 2 Let $ z\in\mathbb{C} \backslash [0,\infty)$ and $Re(s) > 1$. Then $\displaystyle \lim_{\delta \to 0^+}\int_{|z|=\delta}\frac{-z^{s-1}}{e^z -1}\, dz =0$.

Proof - Lemma 2 Fix $s$. Let $A = | Re(s) - 1|$ and let $B = Im(s)$. We use the ML Inequality to obtain an upper bound for our integral. We also assume that $\delta < 1/2$ and apply lemma 1: \begin{equation*} \int_{|z|=\delta}\frac{-z^{s-1}}{e^z -1}\, dz \leq 2\pi\delta \cdot \max_{|z| = \delta} \left| \frac{-z^{s-1}}{e^z -1} \right| \leq 2\pi\delta \cdot \max_{|z| = \delta} \left| \frac{-z^{s-1}}{\delta /2} \right| = 4\pi \cdot \max_{|z| = \delta} \left|-z^{s-1}\right|. \end{equation*} So it only remains to show that \begin{equation*} \lim_{\delta \to 0^+} \max_{|z| = \delta} \left|-z^{s-1}\right| = 0. \end{equation*} Using the principal value of the complex logarithm (for argument $\theta$: $-\pi < \theta \leq\pi$), we rewrite $-z = \delta e^{i\theta} = e^{\log(\delta) +i\theta}$, so that \begin{align*} \left|-z^{s-1}\right| = \left| e^{ \left( \log(\delta) +i\theta \right) [ A + i B] } \right| = \left| e^{A\log(\delta) - B\theta} e^{ i \left( B\log(\delta) + A\theta \right) } \right| = \left| e^{A \log(\delta) - B\theta } \right| = \delta^A e^{- B\theta} \leq \delta^A e^{\pi \left| B \right| }. \end{align*} But $e^{\pi \left| B \right| }$ is fixed. Because $A > 0$, we have $\lim_{\delta \to 0^+} \delta^A = 0$, completing the proof.

End of Proof - Lemma 2

Main Theorem Let $s \in \mathbb{C}$. We recall the traditional definition of $\zeta(s)$: \begin{align*} \zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} = \prod_{p}\frac{1}{ \left( 1-p^{-s} \right)} \quad\text{(where $p$ ranges over an ordered list of all the primes)}, \end{align*} convergent for $Re(s) > 1$. For the contour of integration $C$ (a Hankel contour oriented in the positive direction defined above), we have that: \begin{align} \zeta(s)=\frac{\Gamma(1-s)}{ -2 \pi i} \int_C \frac{-z^{s-1}}{e^z - 1} \, dz \end{align} extends $\zeta(s)$ to a meromorphic function on $\mathbb{C}$, holomorphic except for a simple pole at $s = 1$.

Proof - Main Theorem.
Unless otherwise stated, we assume $Re(s) > 1$. We start with the Gamma function: $$ \Gamma(s) = \int_0^\infty e^{-x} x^{(s-1)} \, dx. $$ Next, we replace $x$ by $nt$ in the integral (so that $dx=ndt$): \begin{align*} \Gamma(s) &= \int_0^\infty e^{-nt} (nt)^{(s-1)} n\, dt = \int_0^\infty e^{-nt}n^s t^{(s-1)} \, dt = n^s \int_0^\infty e^{-nt} t^{(s-1)} \, dt \\ n^{-s}\Gamma(s) &=\int_0^\infty e^{-nt} t^{(s-1)} \, dt \\ \frac{1}{n^s} &= \frac{1}{\Gamma(s)}\int_0^\infty e^{-nt} t^{(s-1)} \, dt. \end{align*} Now sum both sides over $n$: \begin{align} \sum_{n=1}^\infty \frac{1}{n^s} &= \frac{1}{\Gamma(s)}\sum_{n=1}^\infty \int_0^\infty e^{-nt} t^{(s-1)} \, dt \\ \zeta(s) &= \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=1}^\infty e^{-nt} t^{(s-1)} \, dt \\ \zeta(s) &= \frac{1}{\Gamma(s)} \int_0^\infty t^{(s-1)}\sum_{n=1}^\infty e^{-nt} \, dt.\tag{1} \end{align}

Because we are assuming $Re(s) > 1$, the interchange of the sum and integral is justified by the uniform convergence of the sum. Notice that the sum over $n$ in the last integral is just a geometric series with the $n=0$ term missing. (We are justified in using a geometric series, because $0 < e^{-t} < 1$). So, we have: $$ \sum_{n=1}^\infty e^{-nt} = \sum_{n=0}^\infty e^{-nt} - e^{-0t} = \sum_{n=0}^\infty (e^{-t})^n - 1= \frac{1}{1-e^t} -1 = \frac{e^t}{1-e^t}=\frac{1}{e^t-1}. $$

But that means we can restate equation (1) (and while we are at it replace $dt$ with the slightly more standard $dx$), and have: $$ \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty x^{(s-1)}\sum_{n=1}^\infty e^{-nx} \, dx = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} \, dx. $$

If we set $\displaystyle I(s)= \int_0^\infty \frac{x^{s-1}}{e^x - 1} \, dx$, then we have: $\displaystyle \zeta(s) = \frac{1}{\Gamma(s)} I(s) $.

Our next goal is to evaluate $I(s)$ in the complex plane and see where that takes us. Riemann starts with a related but not identical integral, which we call $I_0(s)$: $$ I_0(s) = \int_C \frac{-z^{s-1}}{e^z - 1} \, dz, $$ using the contour of integration $C$ described above. This contour stays well away from the singularities of the integrand at $2\pi i \mathbb{Z}$. We have: $$ I_0(s) = \int_C \frac{-z^{s-1}}{e^z - 1} \, dz = \int_\infty^{p_1} \frac{-z^{s-1}}{e^z - 1} \, dz + \int_{|z|=\delta} \frac{-z^{s-1}}{e^z - 1} \, dz + \int_{p_2}^\infty \frac{-z^{s-1}}{e^z - 1} \, dz. $$ We now study $I_0(s)$ as $\delta\rightarrow 0$. By Lemma 2, the second integral can be disregarded so the contour includes only the horizontal lines $\infty\rightarrow p_1$ and $p_2\rightarrow\infty$.

Note that $(-z)^{(s-1)}=e^{(s-1)log(-z)}$ and that $log(-z)=log|z|+iarg(-z)$. We will use $Arg(z)$, the principal value of $arg(z)$, defined as the value $\theta$ satisfying $-\pi < \theta \leq \pi$. Recall that $z \mapsto Arg(z)$ is discontinuous at each point on the nonpositive real axis. Let $z=x_0+iy$ for some fixed $x_0 < 0$. If $y\downarrow 0$ then $Arg(z) \rightarrow \pi$, whereas, if $y \uparrow 0$ then $Arg(z) \rightarrow -\pi$.
In the first integral, $-z=-\delta_x - i\delta_y$ approaches the nonpositive real axis from below, so (as $\delta\rightarrow 0$) we have $Arg(-z) \rightarrow -\pi$. In the third integral, $-z=-\delta_x + i\delta_y$ approaches the nonpositive real axis from above, so we have $Arg(-z) \rightarrow \pi$. \begin{align*} I_0(s) &= %\int_C \frac{-z^{s-1}}{e^z - 1} \, dz = \int_{\infty}^{p_1} \frac{e^{(s-1)(\log |z| -\pi i)}}{e^z - 1} \, dz + \int_{p_2}^{\infty} \frac{e^{(s-1)(\log |z| +\pi i)}}{e^z - 1} \, dz \\ &= -\int_{p_1}^{\infty} \frac{ e^{(s-1) \log |z| } e^{(s-1)(-\pi i)} }{e^z - 1} \, dz + \int_{p_2}^{\infty} \frac{ e^{(s-1) \log |z| } e^{(s-1)(\pi i)} }{e^z - 1} \, dz \\ &= -e^{(s-1)(-\pi i)} \int_{p_1}^{\infty} \frac{ e^{(s-1) \log |z| } }{e^z - 1} \, dz + e^{(s-1)(\pi i)} \int_{p_2}^{\infty} \frac{ e^{(s-1) \log |z| } }{e^z - 1} \, dz. \end{align*} Taken to the limit, our integrals are over a horizontal line with $dy = 0$, so we can replace $dz$ with $dx$ (and $z$ with $x$), replace $p_1$ and $p_2$ with $0$, and consider $e^{(s-1) \log |z|} = x^{(s-1)}$: \begin{align*} \lim_{\delta \to 0+} I_0(s) &=-e^{(s-1)(-\pi i)} \int_{0}^{\infty} \frac{ e^{(s-1) \log |z| } }{e^z - 1} \, dz + e^{(s-1)(\pi i)} \int_{0}^{\infty} \frac{ e^{(s-1) \log |z| } }{e^z - 1} \, dz \\ &= \left[ e^{i\pi(s-1)} - e^{-i\pi(s-1)} \right] \int_0^\infty \frac{x^{(s-1)}}{e^x - 1} \, dx. \end{align*} But we have: $e^{iw}-e^{-iw} = \cos(w) + i \sin(w) - \left[ \cos(w)-i\sin(w) \right] = 2i\sin(w)$. So, the value in brackets is: $2i\sin(\pi (s-1)) = 2i\sin(\pi s - \pi) = -2i\sin(\pi s)$. This gives our value for $I_0(s)$: \begin{equation*} \lim_{\delta \to 0+} I_0(s) = -2i\sin(\pi s)\int_0^\infty \frac{x^{(s-1)}}{e^x - 1} \, dx = -2i\sin(\pi s) I(s). \end{equation*} Reviewing, we have: \begin{equation*} \zeta(s)= \frac{1}{\Gamma(s)} I(s)\quad\text{and}\quad I_0(s) = -2i\sin(\pi s) I(s) \quad\text{so that}\quad \zeta(s)=\frac{1}{\Gamma(s)}\cdot\frac{1}{ -2i\sin(\pi s) } I_0(s). \end{equation*} Now use a rearrangement of Euler's Reflection Formula: $\displaystyle \frac{1}{\Gamma(s)} = \frac{\Gamma(1-s)\sin(\pi s)}{\pi}$, giving

\begin{equation*} \zeta(s)= \left[ \frac{1}{\Gamma(s)} \right] \frac{1}{ -2i\sin(\pi s) } I_0(s) = \left[ \frac{\Gamma(1-s)\sin(\pi s)}{\pi} \right] \frac{1}{ -2i\sin(\pi s) } I_0(s) = \frac{\Gamma(1-s)}{ -2 \pi i} I_0(s). \end{equation*} We therefore have: \begin{align} \zeta(s)=\frac{\Gamma(1-s)}{ -2 \pi i} \int_C \frac{-z^{s-1}}{e^z - 1} \, dz. \end{align}

Up to this point, we have assumed $Re(s) > 1$. But the integral is an entire function (it is uniformly convergent in any compact subset of $\mathbb{C}$ because $e^z$ grows faster than any power of $z$). That means that $\zeta(s)$ is analytic, except possibly at the positive integers where $\Gamma(1-s)$ has simple poles. But we already know that $\zeta(s)$ is analytic for $Re(s) > 1$, so the possible poles at $2,3,4...$ must be removable and cancel against zeros of the integral. Thus, the only pole of $\zeta(s)$ is at $s= 1$.

End of Proof - Main Theorem.