My understanding is that if this integral exists in the real sense, i.e. real Riemann-wise, then I can apply the residue theorem. If not, I may use the Cauchy Principal Value, to obtain a value.
To begin with the residues, there are two isolated poles: $z = 0$ of order $2$ and $z=k\pi$ of order $1$; $k\in \mathbb{Z}\setminus\{0\}$.
Evaluating the pole at $0$, I expanded $f$ into its series centred at $0$ and found its first negative power coefficient to be $0$.
Then for the residues, I used the residue-pole formula:
$$\lim_{z \to k\pi} \frac{z-k\pi}{z\sin z} = \frac{1}{k\pi \cos(k\pi)}=
\frac{1}{k\pi}.
$$
The sum of these residues diverges by comparing with the harmonic series.
However, integrating on $|z|=5$ I may only consider the $\pi$ pole, since all others lie outside the circle.
Then the integral should be $\frac{2i\pi}{\pi}$=$2i$ using the residue theorem.
However, I don't think I can use these residues to compute an integral that does not exist. Is this correct? I ask this question, because it seemed to be implied that I should compute the integral this way on an exam.
If I cannot use the residues, then I will try using the PV and update this post with the attempt.
Best Answer
The poles of $1/(z \sin{z})$ are at $0$, and $\pm \pi$ within $|z| \lt 5$. The residue at $0$ is $0$ - this should be apparent because $z \sin{z}$ is even. The residue at each of $\pm \pi$ is given by $(\pm 1/\pi)/(\cos{(\pm \pi)}) = \mp 1/\pi$. Thus, the integral is zero.