I was reading Riemann's Zeta Function by H. Edwards, and could not understand the equation on the page 12.

\begin{align*} \zeta(-n) &= \frac{\prod(n)}{2\pi i}\int_{+\infty}^{+\infty} \frac{(-x)^{-n}}{e^x-1}\frac{dx}{x} \\ &= \frac{\prod(n)}{2\pi i}\int_{|x|=\delta} \left( \sum_{m=0}^\infty \frac{B_mx^m}{m!}\right)\frac{(-x)^{-n}}{x}\cdot \frac{dx}{x} \\ &= \sum_{m=0}^\infty \prod(n) \frac{B_m}{m!}(-1)^n \cdot \frac{1}{2\pi} \int_0^{2\pi}x^{m-n-1}d\theta \\ &= n! \frac{B_{n+1}}{(n+1)!}(-1)^n = (-1)^n \frac{B_{n+1}}{n+1}. \end{align*}

Three to note.

  1. $n$ is natural number.

  2. $(s-1)! = \prod(s-1) = \Gamma(s)$.

  3. The contour is, from the positive infinity decrease to $\delta > 0$ along the positive $x$-axis, rotate counter clockwise around zero with the radius of $\delta$, and go back to infinity along the $x$-axis. (It seems to take $\delta \to 0$.)

I do not understand three parts from these equations.

  1. How was the integral in the first line simplified into the second one?

  2. Why was the integral in the third line disappeared?

  3. Which recurrence relation of Bernoulli number was used here?

  • $\begingroup$ It is known that the taylor expansion around zero of $\frac{1}{e^x-1}$ can be expressed in terms of Bernoulli numbers as the author has done. It is a classical exercise in complex analysis. The vanishing of the integral is nothing but the fact that $\int_{0}^{2\pi}e^{inx}dx=0$ for all $n\neq 0$ and equals $2\pi$ if $n=0$, as the primitive function is $2\pi$ periodic. $\endgroup$ – b00n heT Jun 23 '16 at 13:41
  • $\begingroup$ Latex note. Write $\Pi(s-1) = \Gamma(s)$ not $\prod(s-1) = \Gamma(s)$ $\endgroup$ – GEdgar Jun 23 '16 at 14:31
  • $\begingroup$ and for proving it is related to $\zeta(-n)$ see fr.wikipedia.org/wiki/… (demonstration) $\endgroup$ – reuns Jun 23 '16 at 17:56

Expanding my comment:

1) To prove that $$\frac{1}{e^x-1}= \frac1x\cdot\sum_{k\ge 0}\frac{B_n}{n!}x^n$$ you just need to write $\frac{x}{e^x-1}=\sum_{k\ge 0}a_nx^n$, multiply it with the Taylor expansion of $e^x-1$, compare the coefficients, and show that these satisfy the relations satisfied by the Bernoulli numbers.

2) To go from one integral to the next, you need to use the residue theorem on the following indented contour as $R\nearrow \infty$, noticing that the integral over $\Gamma_R$ goes to zero by the classical $ML$ estimate.

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3) As said in the comment, this is just a consequence of $$\int_0^{2\pi}e^{nit}dt=\frac{1}{ni}e^{int}\Big|^{2\pi}_0=0,$$ for $n\neq 0$, and $$\int_0^{2\pi}e^{0}dt=\int_0^{2\pi}1dt=2\pi$$

  • $\begingroup$ (Most of us use $\sum_{k=0}^\infty \frac{B_k}{k!} x^k = \frac{x}{e^x-1}$ as the definition of Bernouilli numbers and then deduce their properties/recurrence, at least in the theory of $\zeta(s)$) $\endgroup$ – reuns Nov 14 '18 at 15:39
  • $\begingroup$ Good to know @reuns. I had actually never seen them until I saw them in this exercise. What field do you specialise in, that uses these? $\endgroup$ – b00n heT Nov 14 '18 at 17:17

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