Proving that $ \lim_{R \rightarrow +\infty} \int_{\gamma_{R_{N}}} \frac{iRe^{i\theta}d\theta}{R^4e^{4i\theta}\sin(\pi Re^{i\theta}) } = 0 $ I want to show that:
$$ S = \sum_{n=1}^{\infty} \frac {(-1)^{n}}{n^4} = - \frac {7\pi^4}{720} $$
In order to do that, I use the residue theorem method. I define the function:
$$ f(z) = \frac{1}{z^4\sin(\pi z)} $$
The poles are $0$ and $ \left\{z = n, n \in \mathbb{Z^{*}} \right\}$.
The residues are $$\text{Res}(f,0) = \frac{7\pi^3}{360}$$ and 
$$\text{Res}(f,n) = \frac{(-1)^n}{\pi n^4}, \qquad\forall n \in \mathbb{Z^{*}}.$$
I then define $\gamma_{R_N}$ to be the circle with center $z=0$ and radius $N<R<N+1$ for $N\in\mathbb{N}^*$, i.e., 
$$\gamma_{R_{N}} = \left\{z = Re^{i \theta}, \theta \in [0,2\pi] \right\}.$$
In order to use the residue theorem, I want to proove that:
$$ \lim_{R \rightarrow +\infty} \int_{\gamma_{R_{N}}} \frac{dz}{z^4\sin(\pi z) } = 0 $$ 
Here $z = Re^{i\theta} $ so I have to prove that:
$$ \lim_{R \rightarrow +\infty} \int_{\gamma_{R_{N}}} \frac{iRe^{i\theta}d\theta}{R^4e^{4i\theta}\sin(\pi Re^{i\theta}) } = 0 $$ 
How can I do that? 
Thank you all!
 A: This is not a complete solution but I'll post it anyway; maybe you'll find it useful 
Bound the modulus of the integral by something that tends to $0$ faster than $1/R$. If the modules of the integral is bounded by something that tends to $0$, the integral too will tend to $0$
In this case, show that $$|\sin(\pi z)| \ge \frac{e^{\pi R \sin \theta} - e^{-\pi R \sin \theta}}2$$ 
Hence you can write 
$$\left | \int _{\gamma_R} \frac 1{z^4 \sin (\pi z)} \right | \le \int _{\gamma_R} \frac 2{R^4 (e^{\pi R \sin \theta} - e^{-\pi R \sin \theta})} dz $$
Unfortunaltey, the integrand blows up when $\theta = 0, \pi$ (i.e. when the sine is $0$). On the other hand, there are the only two points where there can be a problem; so define $$\tilde f(z) = \begin{cases} \frac 1{z^4 \sin (\pi z)} & \text{$\arg z = \theta \notin \{0, \pi\}$} \\ 0  & \text{otherwise}\end{cases}$$
Since $f(z) = \tilde f(z)$ almost everywhere, you have that 
$$\int _{\gamma_R} \frac 1{z^4 \sin (\pi z)}dz= \int _{\gamma_R} \tilde f(z) dz$$
You can then bound the modulus of the latter integral (much in the same way as before) to find 
$$ \left|\int _{\gamma_R} \tilde f(z) dz\right| \le \int _{\gamma_R} \frac 2{R^4 (e^{\pi R \sin \theta} - e^{-\pi R \sin \theta})} (1 - 1_{\{0, \pi\}}(\theta)) dz = \int_0^{2\pi} \frac {2(1 - 1_{\{0, \pi\}}(\theta))}{R^3 (e^{\pi R \sin \theta} - e^{-\pi R \sin \theta})} d\theta$$
Taking the limit as $R \to \infty$, if you can exchange the limit and the integral you would find the result you want i.e. the latter integral  tends to $0$, hence also the integral of $\tilde f(z)$ tends to $0$. hence also the integral of $f(z)$ tends to $0$.  Not sure how to prove that it's possible to exchange the limit and the integral though; lebesgue dominated convergence theorem doesn't help since the function is not bounded by something integrable. Not sure how to proceed; it may be that the integral doesn't go to $0$
