Contour Integration, Riemann Zeta (-n)

Question

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?

Answer 1

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.

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 $$\int_0^{2\pi}e^{0}dt=\int_0^{2\pi}1dt=2\pi$$