= Solution
For $\operatorname{Re}s>1$, expand $(e^t-1)^{-1}=\sum_{n\geq1}e^{-nt}$. <Absolute convergence> of the integrated series follows from the same expression with $\operatorname{Re}s$ in place of $s$. Integrating termwise and substituting $u=nt$ gives
$$
\int_0^\infty\frac{t^{s-1}}{e^t-1}\,dt=\sum_{n\geq1}n^{-s}\Gamma(s)=\Gamma(s)\zeta(s).
$$
This proves the <Mellin transform> representation of the <Riemann zeta function>.
Take the <Hankel contour> along the lower side of the negative axis towards zero, around zero counterclockwise, and back along the upper side. For $\operatorname{Re}s>1$ the small-circle contribution tends to zero. On the lower and upper rays, respectively, $\arg t=-\pi,+\pi$, so the two integrals combine to
$$
\int_H\frac{t^{s-1}}{e^{-t}-1}\,dt=\left(e^{-i\pi(s-1)}-e^{i\pi(s-1)}\right)\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx=2i\sin(\pi s)\Gamma(s)\zeta(s).
$$
The <gamma reflection formula> now proves agreement of the two representations. At an integer $s=m\geq2$, the contour integral has a zero while $\Gamma(1-s)$ has a pole: \b[their product is interpreted by its finite limit, equal to $\zeta(m)$], not by multiplying literal zero and infinity.
For nonpositive integers the integrand is single-valued, and the contour closes around zero. Since
$$
\frac1{e^{-t}-1}=-\frac1t-\frac12+O(t),
$$
the <residue> of $t^{-1}/(e^{-t}-1)$ is $-1/2$, giving \b[$\zeta(0)=-1/2$]. Also
$$
\frac1{e^{-t}-1}+\frac12=-\frac12\coth(t/2)
$$
is odd. Its Laurent series therefore has only odd powers. Multiplication by $t^{-2n-1}$ cannot produce a $t^{-1}$ term, and the subtracted constant cannot do so when $n\geq1$. Hence the contour integral vanishes and \b[$\zeta(-2n)=0$ for every positive integer $n$].
Back to article page
