Example 4. Compute the integral

$$
\int_{|x|=2} \frac{1}{z-1} \sin \frac{1}{z} d z
$$
Solution. In the circle $|z| \leqslant 2$, the integrand has two singular points $z=1$ and $z=0$. It is easy to establish that $z=1$ is a simple pole, therefore

$$
\operatorname{res}\left(\frac{1}{z-1} \sin \frac{1}{z}\right)=\left.\frac{\sin \frac{1}{z}}{(z-1)^{\prime}}\right|_{z=1}=\sin 1
$$

To determine the nature of the singular point $z=0$, we write the Laurent series for the function $\frac{1}{z-1} \sin \frac{1}{z}$ in the neighborhood of this point. We have

$$
\begin{aligned}
\frac{1}{z-1} \sin \frac{1}{z}= & -\frac{1}{1-z} \sin \frac{1}{z}=-\left(1+z+z^{2}+\ldots\right)\left(\frac{1}{z}-\frac{1}{3!z^{3}}+\frac{1}{5!z^{5}}-\ldots\right)= \\
= & -\left(1-\frac{1}{31^{\prime}}+\frac{1}{5!}-\ldots\right) \frac{1}{z}+\frac{c_{-2}}{z^{2}}+\frac{c_{-3}}{z^{3}}+\ldots+\text { regular part } \\
& c_{-k} \neq 0, \quad k=2,3, \ldots
\end{aligned}
$$

Since the Laurent series contains an infinite number of terms with negative powers of $z$, the point $z=0$ is an essential singularity. The residue of the integrand at this point is

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-089.jpg?height=131&width=706&top_left_y=685&top_left_x=349)

Therefore,

$$
\int_{|z| \neq 2} \frac{1}{z-1} \sin \frac{1}{z} d z=2 \pi i(\sin 1-\sin 1)=0
$$

## Problems for Independent Solution

Calculate the integrals:

347. $\int_{|z|=1} z \operatorname{tg} \pi z d z . \quad$ 348. $\int_{C} \frac{z d z}{(z-1)^{2}(z+2)}$, where $C: x^{2 / 3}+y^{2 / 3}=3^{2 / 3}$,
348. $\int_{\mid z==2} \frac{e^{z} d z}{z^{3}(z+1)}$
349. $\int_{|z-i|=3} \frac{e^{z^{2}}-1}{z^{3}-i z^{2}} d z . \quad 351 . \quad \int_{|z|=1 / 2} z^{2} \sin \frac{1}{z} d z$.
350. $\int_{|z|=\sqrt{3}} \frac{\sin \pi z}{z^{2}-z} d z$
351. $\int_{|z+1|=4} \frac{z d z}{e^{z}+3}$
352. $\int_{|z|=1} \frac{z^{2} d z}{\sin ^{3} z \cos z}$
353. $\int_{|z-i|=1} \frac{e^{z} d z}{z^{4}+2 z^{2}+1} \cdot 356 . \int_{|x|=-4} \frac{e^{i z} d z}{(z-n)^{3}}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-089.jpg?height=138&width=1131&top_left_y=1689&top_left_x=136)
359. $\int_{C} \frac{\sin \pi z}{\left(z^{2}-1\right)^{2}} d z, C: \frac{x^{2}}{4}+y^{2}=1 . \quad 360 . \int_{C} \frac{z+1}{z^{2}+2 z-3} d z, \quad C: x^{2}+y^{2}=16$.

361. $\int_{C} \frac{z \sin z}{(z-1)^{5}} d z, \quad C: \frac{x^{2}}{3}+\frac{y^{2}}{9}=1$. 362. $\int_{C} \frac{d z}{z^{4}+1} d z, \quad C: x^{2}+y^{2}=2 x$.
362. $\int_{|z|=1} z^{3} \sin \frac{1}{z} d z . \quad$ 364. $\int_{|z|=1 / 3}(z+1) e^{1 / z} d z$.
363. $\int_{|z|=2 / 3}\left(\sin \frac{1}{z^{2}}+e^{z^{2}} \cos z\right) d z$.

Residue of a function at the point at infinity

It is said that the function $f(z)$ is analytic at the point at infinity $z=\infty$, if the function

$$
\varphi(\zeta)=f\left(\frac{1}{\zeta}\right)
$$

is analytic at the point $\zeta=0$.

For example, the function $f(z)=\sin \frac{1}{z}$ is analytic at the point $z=\infty$, since the function

$$
\varphi(\zeta)=f\left(\frac{1}{\zeta}\right)=\sin \zeta
$$

is analytic at the point $\zeta=0$.

Let the function $f(z)$ be analytic in some neighborhood of the point at infinity (except the point $z=\infty$ itself).

The point $z=\infty$ is called an isolated singular point of the function $f(z)$ if there are no other singular points of the function $f(z)$ in some neighborhood of this point.

The function $f(z)=\frac{1}{\sin z}$ has a non-isolated singularity at infinity: the poles $z_{k}=k \pi$ of this function accumulate at infinity as $k \rightarrow \infty$.

It is said that $z=\infty$ is a removable singularity, a pole, or an essential singularity of the function $f(z)$ depending on whether $\lim _{z \rightarrow \infty} f(z)$ is finite, infinite, or does not exist.

The criteria for the type of singularity at infinity, related to the Laurent series expansion, differ from the criteria for finite singular points.

Theorem 1. If $z=\infty$ is a removable singularity of the function $f(z)$, then the Laurent series expansion of $f(z)$ in the neighborhood of this point does not contain positive powers of $z$; if $z=\infty$ is a pole, then this expansion contains a finite number of positive powers of $z$; in the case of an essential singularity - an infinite number of positive powers of $z$.

The Laurent series expansion of the function $f(z)$ in the neighborhood of the point at infinity is called the Laurent series expansion of $f(z)$ that converges everywhere outside a circle of sufficiently large radius $R$ centered at the point $z=0$ (except, perhaps, the point $z=\infty$ itself).

Let the function $f(z)$ be analytic in some neighborhood of the point $z=\infty$ (except, perhaps, the point $z=\infty$ itself).

The residue of the function $f(z)$ at infinity is defined as

$$
\operatorname{res} f(\infty)=\frac{1}{2 \pi i} \int_{\gamma^{-}} f(z) d z
$$

where $\gamma^{-}$ is a sufficiently large circle $|z|=\rho$, traversed clockwise (so that the neighborhood of the point $z=\infty$ remains on the left, as in the case of a finite point $z=a$).

From this definition, it follows that the residue of the function at infinity is equal to the coefficient of $z^{-1}$ in the Laurent series expansion of $f(z)$ in the neighborhood of $z=\infty$, taken with the opposite sign:

$$
\operatorname{res} f(\infty)=-c_{-1}
$$