Analytic Continuation of the Riemann Zeta Function via Contour Integration

Question

I've read a number of zeta function related texts and questions on forums like this and I still do not understand some parts of how $\zeta$ is analytically (meromorphically) continued using contour integration as Riemann did. I understand that $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$ converges for $\Re(s)>1$ and that Dirichlet's Eta function can be used to analytically continue it to $\Re(s)>0$.

The Hankel contour is often mentioned in relation to this topic. I see it as part of a keyhole contour, $C$, for the function: $$\frac{z^{s-1}}{\mathrm{e}^z-1}$$ Taking the branch cut to lie along the positive real axis, if we orient the keyhole contour positively (anticlockwise) and examine the integral over the Hankel contour $\mathcal{H}$, we get the following:$$\int_{\mathcal{H}} \frac{z^{s-1}}{\mathrm{e}^z-1} \mathrm{d}z = (1-\mathrm{e}^{2\pi is}) \int_0^\infty \frac{x^{s-1}}{\mathrm{e}^x-1} \mathrm{d}x - \int_\gamma \frac{z^{s-1}}{\mathrm{e}^z-1} \mathrm{d}z$$ The two integrals on the RHS converge for $\Re(s)>1$, where the integral over $\gamma$ is the infinitesimal circle around the origin. I get confused around here. If we want to apply the residue theorem, we require a closed contour. A circle of radius $R$ completes the keyhole contour, an integral which only converges for $\Re(s)<0$. Applying the residue theorem gives $2(2\pi i)^s \zeta(1-s)$ as the value of the integral over $C$. This, however, also only converges for $\Re(s)<0$.

We can write an equality by separating the terms into some which converge for $\Re(s)>1$ and others which converge for $\Re(s)<0$. I get that the essence of analytic continuation is to extend the domain of definition but these two regions have no overlap, a condition which is required for an analytic continuation. It makes more sense to me in the context of Eta as we have already got convergence for $\zeta$ when $\Re(s)>0$ however the literature often makes this jump without mention of Eta which leaves me wondering as to how this is justified. This is likely due to a basic misunderstanding on my part of some of the concepts at play here but I appreciate any help that I'm offered.

Answer 1

I have also been wondering for a long time how the integral over Hermann Hankel's contour is supposed to converge for any finite variable $s$. A friend recently helped me with this issue. Here's what's hidden behind the words "trivial/obvious" in many publications.

Starting with Hankel's contour where $\epsilon$ goes to zero and $R$ goes to infinity:

We have for $\Re(s)>1$ that:

$$ \lim_{\epsilon \to 0}\lim_{R \to \infty} \int_{\mathcal{H}_{\epsilon,R}}\frac{z^{s-1}}{e^{z}-1}dz=(1-e^{2\pi is})\int_0^\infty\frac{x^{s-1}}{e^{x}-1}dx $$

But now introducing the modified Hankel's contour:

$$ \int_{\mathcal{H}'_R}\frac{z^{s-1}}{e^{z}-1}dz=-\int_i^{R+i}\frac{z^{s-1}}{e^{z}-1}dz+\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{(e^{i\theta})^{s-1}}{e^{e^{i\theta}}-1}ie^{i\theta}d\theta+\int_{-i}^{R-i}\frac{z^{s-1}}{e^{z}-1}dz $$ Taking a closer look:

$$ \left|\int_{1+i}^{R+i}\frac{z^{s-1}}{e^{z}-1}dz\right|=\left|\int_1^R\frac{(x+i)^{s-1}}{e^{x+i}-1}dx\right|\leqslant\int_1^R\left|\frac{(x+i)^{s-1}}{e^{x+i}-1}\right|dx=\int_1^R\left|\frac{(x+i)^{s-1}e^{-x-i}}{1-e^{-x-i}}\right|dx $$

and for $x\geqslant1$:

$$ \left|{1-e^{-x-i}}\right|\geqslant1-\left|e^{-x-i}\right|=1-e^{-x}\geqslant 1-e^{-1}\geqslant\frac{1}{2} $$

$$ \left|{e^{-x-i}}\right|=e^{-x} $$

$$ \left|{(x+i)^{s-1}}\right|=\left|{e^{(s-1)\ln(x+i)}}\right|=\left|{e^{(s-1)(\frac{1}{2}\ln(x^2+1)+i\arctan(\frac{1}{x}))}}\right|={e^{\frac{1}{2}\Re(s-1)\ln(x^2+1)-\Im(s-1)\arctan(\frac{1}{x})}}\leqslant $$ $$ \leqslant{e^{|\Re(s-1)|\ln(2x^2)+\frac{\pi}{4}|\Im(s-1)|}}=e^{|\Re(s-1)|\ln(2)+2|\Re(s-1)|\ln(x)+\frac{\pi}{4}|\Im(s)|} $$

For $x$ large enough we have (above a certain $X$): $$|\Re(s-1)|\ln(2)+2|\Re(s-1)|\ln(x)+\frac{\pi}{4}|\Im(s)|\leqslant\frac{x}{2}$$

Therefore:

$$ \int_X^R\left|\frac{(x+i)^{s-1}e^{-x-i}}{1-e^{-x-i}}\right|dx\leqslant\int_X^R 2e^{-\frac{x}{2}}dx $$ which is finite and hence:

$$\int_{1+i}^{R+i}\frac{z^{s-1}}{e^{z}-1}dz$$

must also converge regardless of the variable $s$. You can do similar tricks to show that: $$\int_{1-i}^{R-i}\frac{z^{s-1}}{e^{z}-1}dz$$

converge regardless of the variable $s$ too. The rest of the integral over modified Hankel's contour inclouding:

$$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{(e^{i\theta})^{s-1}}{e^{e^{i\theta}}-1}ie^{i\theta}d\theta$$

has finite lenght and does not pass through any irregularities such as singularities and branch cuts. So:

$$\lim_{R \to \infty}\int_{\mathcal{H}'_R}\frac{z^{s-1}}{e^{z}-1}dz=\int_{\mathcal{H}'}\frac{z^{s-1}}{e^{z}-1}dz$$

converges for any $s$.

Combining two contours into a closed contour $\mathcal{C}$:

and using Cauchy's integral theorem, because inside $\mathcal{C}_{\epsilon,R}$ our integrand is holomorphic:

$$ \oint_{\mathcal{C}_{\epsilon,R}}=-\int_{\mathcal{H}_{\epsilon,R}}+\int_{\mathcal{H}_R'}+\int_{R+i\epsilon}^{R+i}+\int_{R-i}^{R-i\epsilon}=0 $$

Estimating:

$$ \left|\int_{R+i\epsilon}^{R+i}\frac{z^{s-1}}{e^{z}-1}dz+\int_{R-i}^{R-i\epsilon}\frac{z^{s-1}}{e^{z}-1}dz\right|=\left|\int_{\epsilon}^{1}\frac{(R+iy)^{s-1}}{e^{R+iy}-1}dy+\int_{-1}^{-\epsilon}\frac{(R+iy)^{s-1}}{e^{R+iy}-1}dy\right|\leqslant $$ $$\text{doing similar things as we did with integral from X to R}$$ $$ \leqslant\int_{-1}^{1}2e^{-R}{e^{\frac{1}{2}|\Re(s-1)|\ln(R^2+y^2)+|\Im(s-1)\arctan(\frac{y^2}{R^2})|}}dy\leqslant\int_{-1}^{1}2e^{-R}{e^{|\Re(s-1)|\ln(2R^2)+\frac{\pi}{4}|\Im(s)|}}dy $$

taking $R \to \infty$ we showed that integrals over red lines from contour $\mathcal{C}$ vanishes.

$$ -\int_{\mathcal{H}_{\epsilon}}+\int_{\mathcal{H}'}=0 $$ and then take $\epsilon \to 0$.

Finally, we have that the integral over the $\mathcal{H}$ agrees to the integral over the modified contour $\mathcal{H'}$ for $\Re(s)>1$. What is most important is that by replacing $\mathcal{H}$ with $\mathcal{H'}$ we analytically continue the integral to region $s\in\mathbb{C}$. And that's why, in the next step, if we add a large circle to $\mathcal{H'}$, the integral over that large circle will converge because we can now choose $\Re(s)<0$.