### 3.34. Compute the integral

$$
\int_{L} \frac{\sin z}{z\left(z-\frac{\pi}{2}\right)} d z
$$

where $L-$ is a rectangle bounded by the following lines: $x=2, x=-1, y=2, y=-1$.
Solution. The special points of the integrand function, which in this example has the form

$$
f(z)=\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}
$$

are the points $z=0$ and $z=\frac{\pi}{2}$. The point $z=0$ is a removable singularity of the integrand function, since at this point

$$
\lim _{z \rightarrow 0} f(z)=\lim _{z \rightarrow 0} \frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}=\lim _{z \rightarrow 0} \frac{\sin z}{z} \lim _{z \rightarrow 0}\left(z-\frac{\pi}{2}\right)=1 \cdot\left(-\frac{\pi}{2}\right)=-\frac{\pi}{2}
$$

Therefore, at this point

$$
\operatorname{res}\left(\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}, 0\right)=0
$$

The point $z=\frac{\pi}{2}$ is a simple pole of the integrand function. Therefore, according to the formula for calculating the residue at a simple pole, we have:

$$
\operatorname{res}\left(\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}, \frac{\pi}{2}\right)=\lim _{z \rightarrow \frac{\pi}{2}}\left(z-\frac{\pi}{2}\right) \cdot \frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}=\lim _{z \rightarrow \frac{\pi}{2}} \frac{\sin z}{z}=\frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}
$$

Now, by the main theorem of residues, we get

$$
\begin{aligned}
& \int_{L} \frac{\sin z}{z\left(z-\frac{\pi}{2}\right)} d z=2 \pi i\left(\operatorname{res}\left(\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}, 0\right)+\operatorname{res}\left(\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}, \frac{\pi}{2}\right)\right)= \\
& 3.35 \text { Find } \quad=2 \pi i i\left(0+\frac{2}{\pi}\right)=4 i .
\end{aligned}
$$

$$
\int_{L} \sin \frac{i}{z} d z
$$

where $L-$ is a circle of radius $r>0$, i.e., $|z|=r$.

Solution. In the region of the complex plane bounded by the specified circle, the integrand function $f(z)=\sin \frac{i}{z}$ has a single singular point $z=0$, which is an essential singularity of this function. Indeed, the Laurent series expansion of this function in the annulus $0<|z|<r$ is

$$
\sin \frac{i}{z}=\sum_{n=0}^{x} \frac{(-1)^{2 n+1}}{(2 n+1)!}\left(\frac{i}{z}\right)^{2 n+1}=\frac{i}{z}-\frac{1}{3!}\left(\frac{i}{z}\right)^{3}+\ldots
$$

From the given Laurent series expansion of the integrand function, it follows that the coefficient $c_{-1}=i$. Therefore, $\operatorname{res}\left(\sin \frac{i}{z}, 0\right)=c_{-1}=i$. According to the main theorem of residue theory, we have:

$$
\int_{L} \sin \frac{i}{z} d z=2 \pi i \operatorname{res}\left(\sin \frac{i}{z}, 0\right)=2 \pi i \cdot i=2 \pi i^{2}=2 \pi(-1)=-2 \pi
$$