analytic continuation of Riemann zeta (using integral)

The Riemann zeta function can be analyticallycontinued to the whole complex plane minus the point 1by means of an integral representation. Remember that the zeta functonis defined by the series

$$\zeta (s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}.$$

When $\mathrm{\Re }s>1$, this series converges; furthermore, this convergence isuniform on compact subsets of this half-plane, hence the series converges toan analytic function on this half plane. However, the series diverges whenwe have , so this series cannot be used to define the zeta functionin the whole complex plane, which is why we must make an analytic continuation.

To make this continuation, we start by changing the variable in an integration:

$$n^s{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{s-1}𝑑x={\int }_0^{\mathrm{\infty }}{e}^{-y}{y}^{s-1}𝑑y=\mathrm{\Gamma }(s)$$

This provides us with an integral representation of our summand.Substituting this into the series, we find that

$$\zeta (s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{s-1}𝑑x=\frac{1}{\mathrm{\Gamma }(s)}\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{s-1}𝑑x.$$

We note that

$$\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\left|e^{-nx}{x}^{s-1}\right|𝑑x=\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{|s-1|}𝑑x=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(|s-1|+1)}{n^s};$$

because the series converges, hence it is possibleto interchange integration and summation andsubsequently sum a geometric series.

$$\zeta (s)=\frac{1}{\mathrm{\Gamma }(s)}\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{s-1}𝑑x=\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{e}^{-nx}{x}^{s-1}dx=\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}𝑑x$$

As it stands, the integral representation we have is not ofmuch use for analytically continuing the zeta functionbecause the integral diverges when on account ofthe fact that the integrand behaves like $x^{-s}$ when $x$is close to zero. However, it is possible to make use of thetheorem of Cauchy to move the path of integration away fromzero.

Given a real number $r>0$, define the contour $C_r$ on theRiemann surface of $z^{s-1}$ as follows: $C_r$ passes from$+\mathrm{\infty }$ to $r$ along a lift of the real axis, then continuesalong the circle of radius $r$ clockwise, and finally goesfrom $r$ to $+\mathrm{\infty }$.

We now examine the integral over such a contour by breaking itinto three pieces.

$${\int }_{C_r}\frac{x^{s-1}}{e^x-1}𝑑x={\int }_r^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}𝑑x+{\int }_{|x|=r}\frac{x^{s-1}}{e^x-1}𝑑x-{\int }_r^{\mathrm{\infty }}\frac{{(e^{-2\pi i}x)}^{s-1}}{e^x-1}𝑑x.$$

We may estimate the third integral in absolute value like so:

$$\left|{\int }_{|x|=r}\frac{x^{s-1}}{e^x-1}𝑑x\right|\le 2\pi r\underset{|x|=r}{sup}\left|\frac{x^{s-1}}{e^x-1}\right|$$

The expression $x/(e^x-1)$ represents an analytic function of$x$, and hence a bounded function of $x$ in a neighborhood of $0$.When $\mathrm{\Re }s>1$, it happens that ${lim}_{x\to 0}|x|x^{s-2}=0$, so

$$\underset{r\to 0}{lim}\left|{\int }_{|x|=r}\frac{x^{s-1}}{e^x-1}𝑑x\right|=0.$$

The third integral differs from the first integral by a phase, sothey may be combined by pulling out this common factor. When$\mathrm{\Re }s>0$, we may take the limit as $r$ approaches $0$ after doingso to obtain the following:

$$\underset{r\to 0}{lim}{\int }_{C_r}\frac{x^{s-1}}{e^x-1}𝑑x=\left(1+e^{2\pi i(1-s)}\right){\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}𝑑x$$

Since, aside from the branch point at$0$, the only singularities of our integrand occur at multiplesof $2\pi i$, it follows from Cauchy’s theorem that

$${\int }_{C_a}\frac{x^{s-1}}{e^x-1}𝑑x={\int }_{C_b}\frac{x^{s-1}}{e^x-1}𝑑x$$

whenever and , which trivially implies that

$$\underset{r\to 0}{lim}{\int }_{C_r}\frac{x^{s-1}}{e^x-1}𝑑x={\int }_{C_r}\frac{x^{s-1}}{e^x-1}𝑑x$$

for any $r$ between $0$ and $2\pi$. Therefore,

$$\zeta (s)=\frac{1}{\left(1+e^{2\pi i(1-s)}\right)\mathrm{\Gamma }(s)}{\int }_{C_{\pi }}\frac{x^{s-1}}{e^x-1}𝑑x$$

when $\mathrm{\Re }z>1$. This integral converges for all complex $s$ becausethe exponential grows more rapidly than the power. Furthermore, thisintegral defines an analytic function of $s$, so we have an analyticcontinuation of the zeta function to the whole complex plane minus the point 1.