In one of my textbook the following problem is written:
Proving the functional equation for the $\zeta$-function:
$$\zeta(z) = 2^z\pi^{z-1}\sin\frac{\pi z}{2} \Gamma(1-z)\zeta(1-z) \qquad \forall z \in \mathbb{C}\setminus \mathbb{N}$$
The proof uses the residue-theorem over the following contour and the function $$g(w) = \frac{w^{z-1}}{e^w-1}$$ where $z$ is the the argument of the $\zeta$-function.
In the proof there is written - Rudin-style ;)
Because $\int_{\color{blue}{\gamma'_n}} g(w) \text{d}w \to 0$ as $n\to \infty$ (left as exercise) $\ldots$
And in the exercises there is as expected:
If $a=\operatorname{Re} z> 1 \quad (z=a+bi)$
Show $|e^w-1| \geqslant \frac{1}{2}$ on $\color{blue}{\gamma'_n}$. Derive from this result $\int_{\color{blue}{\gamma'_n}} g(w) \text{d}w \to 0$ as $n\to \infty$.
I've managed to show $\forall w \in \color{blue}{\gamma'_n}: |e^w-1| \geqslant \frac{1}{2}$.
But how do you show the second part?
$$\left| \int_{\color{blue}{\gamma'_n}} g(w) \text{d}w\right| \leqslant 2\int_{\color{blue}{\gamma'_n}} |w^{z-1}| |\text{d}w |$$
For instance
(Caution for the branch cut)
Using $|w^{z-1}| = |w|^{a-1}e^{-b\arg w}$ if $\operatorname{Im} w>0$ and $\Gamma_1 \leftrightarrow w(y) = n+ iy$ where $y:i\to (2n+1)\pi$.
$$\ldots \leqslant 2\int_{1}^{(2n+1)\pi} |n+iy|^{a-1} e^{—b\arg (n+iy)} d y$$
But why would this $\to 0$ as $n\to \infty$? I know that as $n\to \infty$, $e^{—b\arg(n+iy)}$ can be bounded by a constant. (according to the sign of $b$) But is $|n+iy|^{a-1}$ unbounded as $n\to \infty$.
(Riemann must have been unbelievably genious)
Yes, this is one of Riemanns original proofs. Here's an idea. Since the branch cut for $\log$ is along the positive real axis, we may assume $0\le b<2\pi$. So, $|w^{z-1}|\le C|w|^{a-1}$ or $O(|w|^{a-1})$ if you know that notation. Since $|w|=|-w|$, the integration of $|w|^{a-1}$ along the vertical sides should cancel out.