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In my complex analysis textbook by Stein and Shakarchi, as an exercise, I am supposed to extend $\zeta(s)$ to the entire complex plane using Bernoulli numbers, but I am stuck.

I can prove that

$$ \zeta(s) = \frac{1}{\Gamma(s)} \int_0^1 \frac{x^{s-1}}{e^{x}-1} dx + \frac{1}{\Gamma(s)} \int_1^\infty \frac{x^{s-1}}{e^{x}-1} dx $$

I can further prove that the second integral is an entire function and that given the generating function for the Bernoulli Numbers:

$$ \frac{x}{e^x -1} = \sum_{m=0}^\infty \frac{B_m}{m!} x^m $$

that

$$ \int_0^1 \frac{x^{s-1}}{e^x-1}dx = \sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)} $$

It is clear that there is a pole at $s=1$, but beyond that I don't have any idea why the last summation converges for all $s \neq 1$, which would prove the analytic continuation of $\zeta(s)$ to the entire complex plane.

This is a problem in a first year graduate student textbook, so I imagine that there is a short solution. Thanks for the help.

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    $\begingroup$ The sum also has poles at $s = 0, -1, -2,-3, \dotsc$, but those cancel against the zeros of $\frac{1}{\Gamma(s)}$. For the convergence of the sum, note that the series for $\frac{z}{e^z-1}$ has radius of convergence $2\pi$. $\endgroup$ – Daniel Fischer Jul 3 '14 at 20:05
  • $\begingroup$ I think I am still missing something. Because the radius of convergence is 2*pi, I get that the integral of x/[exp(x)-1] would converge, but why does that still hold when we multiply that equation by x^(s-2) on both the left and right hand side? $\endgroup$ – Nick Jul 3 '14 at 23:29
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Since $\frac{z}{e^z-1}$ is holomorphic on the disk $\{ z : \lvert z\rvert < 2\pi\}$, and has poles in $\pm 2\pi i$, the radius of convergence of the series

$$\sum_{m=0}^\infty \frac{B_m}{m!}z^m$$

is $2\pi$. Hence, by the Cauchy-Hadamard formula,

$$\limsup_{m\to\infty} \left(\frac{\lvert B_m\rvert}{m!}\right)^{1/m} = \frac{1}{2\pi},$$

so for every $R < 2\pi$, for all large enough $m$, we have

$$\frac{\lvert B_m\rvert}{m!} < \frac{1}{R^m}.$$

We know a bit more from the representations of the cotangent: for $\mu\geqslant 1$ we have $B_{2\mu+1} = 0$, and

$$\frac{B_{2\mu}}{(2\mu)!} = \frac{2(-1)^{\mu+1}}{(2\pi)^{2\mu}}\zeta(2\mu),$$

so for all $m\geqslant 0$

$$\frac{\lvert B_m\rvert}{m!} < \frac{4}{(2\pi)^m}.$$

Since the series converges uniformly on the interval of integration, we can interchange integration and summation, and

$$\int_0^1 \frac{x^{s-1}}{e^x-1}\,dx = \int_0^1 \sum_{m=0}^\infty \frac{B_m}{m!} x^{m+s-2}\,dx = \sum_{m=0}^\infty \frac{B_m}{m!}\int_0^1 x^{m+s-2}\,dx = \sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)}.$$

For any $K > 0$, if $\lvert s\rvert \leqslant K$ and $m \geqslant K+2$, we have $\lvert s+m-1\rvert \geqslant 1$, so the sum

$$\sum_{m = \lceil K\rceil+2}^\infty \frac{B_m}{m!(s+m-1)}$$

converges uniformly on the disk $\{ s : \lvert s\rvert \leqslant K\}$ by the Weierstraß $M$-test. Hence

$$H(s) = \sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)}$$

is an entire meromorphic function with simple poles in $1-m,\, m\in\mathbb{N}$.

Since $\Gamma(s)$ has simple poles in $-k,\,k\in\mathbb{N}$, and no zeros, the function

$$\frac{H(s)}{\Gamma(s)} = \frac{1}{\Gamma(s)}\sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)}$$

is an entire meromorphic function with only one simple pole, in $1$, with resdiue $\frac{B_0}{0!\Gamma(1)} = 1$.

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  • $\begingroup$ Thank you for this answer. I understand the solution and I have no further questions. $\endgroup$ – Nick Jul 8 '14 at 0:06

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