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By Taylor expanding

$$\frac{x}{e^{x}-1} = \sum_{n=0}^\infty \frac{B_n}{n!}x^n$$

in the Zeta function

$$\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty x^{s-2} (\frac{x}{e^{x}-1})dx$$

we find

\begin{align} \zeta(s) &= \frac{1}{\Gamma(s)} \int_0^\infty x^{s-2} (\frac{x}{e^{x}-1})dx = \frac{1}{\Gamma(s)} \int_0^\infty x^{s-2} \sum_{n=0}^\infty \frac{B_n}{n!}x^n dx \\ &= \frac{1}{\Gamma(s)} \sum_{n=0}^\infty \frac{B_n}{n!} \int_0^\infty x^{s + n - 2} dx \end{align}

what exactly do we do next to arrive at the formulas for $\zeta(-n)$ and $\zeta(2n)$?

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  • $\begingroup$ Your last integral doesn't converge, so that's not going to go anywhere. $\endgroup$ – Chappers May 15 '16 at 12:32
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    $\begingroup$ Your expansion for $\frac{x}{e^x-1}$ is only valid for $\vert x \vert < 2\pi$ ... so that could be a problem $\endgroup$ – πr8 May 15 '16 at 12:37
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By Taylor expanding, for $|x| < 2 \pi$ :

$$\frac{x}{e^{x}-1} = \sum_{n=0}^\infty \frac{B_n}{n!}x^n$$

in the Zeta function, for $Re(s) > 1$ :

$$\zeta(s)\Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^{x}-1}dx = \int_0^a x^{s-2} \frac{x}{e^{x}-1}dx + \int_a^\infty \frac{x^{s-1}}{e^{x}-1}dx$$

we find whenever $0 < a < 2 \pi$ and $Re(s)> 1$, inverting $\sum$ and $\int$ by absolute/monotone/dominated convergence :

\begin{align} \zeta(s)\Gamma(s) &= \int_a^\infty \frac{x^{s-1}}{e^{x}-1}dx + \int_0^a x^{s-2} \sum_{n=0}^\infty \frac{B_n}{n!}x^n dx \\ &= \int_a^\infty \frac{x^{s-1}}{e^{x}-1}dx + \sum_{n=0}^\infty\frac{B_n}{n!} \int_0^a x^{s + n - 2} dx\\ &= \int_a^\infty \frac{x^{s-1}}{e^{x}-1}dx + \sum_{n=0}^\infty\frac{B_n}{n!} \frac{a^{s+n-1}}{s+n-1}\end{align}

note how by analytic continuation it stays valid for every $s \in \mathbb{C}$ except at the poles,

and how it tells us the residue of $\zeta(s) \Gamma(s)$ at its poles, and hence the value of $\zeta(-k)$ for $k \in \mathbb{N}$

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