Evaluating $\int_c\frac{1}{\sin\frac{1}{z}}\text{d}z$ over $C= \{z\big\vert|z|=\frac{1}{5}\}$

Question

Evaluate

$$\int\limits_{|z|=\frac{1}{5}} \frac{1}{\sin\frac{1}{z}}\text{d}z$$

My attempt: I know that this function has non isolated singularity at $0$, and simple poles at $\frac{1}{n \pi}$. Which ever methods I know are applicable only when all the singularities are isolated. So I tried with Residue theorem but then I realised that the theorem is not applicable here. So now I am stuck. Can any one help please?

I can observe that there are only two poles outside the curve c, one at $1/\pi$ and the other at infinity. Can this be used?

Thanks in advance.

Answer 1

Change variable to $w = \frac1z$, we have

$$\oint_{C_{1/5}} \frac{1}{\sin\frac1z} dz = \oint_{-C_5} \frac{1}{\sin w} \left(-\frac{dw}{w^2}\right) = \oint_{C_5} \frac{1}{w^2\sin w} dw$$

where $C_R$ and $-C_R$ are the circular contour $|z| = R$ in counterclockwise and clockwise orientation.

There are $3$ poles for $w$ inside $C_5$. A triple pole at $w = 0$ and two simple pole at $w = \pm \pi$.

• For $w \sim 0$, we have

$$\frac{1}{w^2\sin w} \approx \frac{1}{w^2(w - \frac{w^3}{6} + O(w^5))} = \frac{1}{w^3}\left(1 + \frac{w^2}{6} + O(w^4)\right)$$

• For $w \sim \pm \pi$, we have

$$\frac{1}{w^2\sin w} \sim \frac{1}{\pi^2\cos\pi(w \mp \pi)} = -\frac{1}{\pi^2(w \mp \pi)}$$ Summing the residues from these $3$ poles, we get

$$\oint_{C_{1/5}} \frac{1}{\sin\frac1z} dz = 2\pi i\left( \frac{1}{6} - \frac{2}{\pi^2}\right) = \left(\frac{\pi}{3}-\frac{4}{\pi}\right) i \approx -0.2260419935385651 i$$

As a double check, we can compute the integral numerically on wolfram alpha using the command Int[(i/5)/Sin[5*Exp[-i*t]]*Exp[i*t],{t,0,2*Pi}] and the result do match.

Answer 2

$$f(z) = \frac1{\sin{\left ( \frac1{\zeta+\frac1{n \pi}} \right )}} $$

where $\zeta = z-\frac1{n \pi}$. Then

$$\begin{align}f(z) &= \frac1{\sin{\left ( \frac{n \pi}{1+ n \pi \zeta} \right )}} \\ &= \frac1{\sin{\left [ n \pi \left (1-n \pi \zeta + n^2 \pi^2 \zeta^2-\cdots \right )\right ]}}\\ &= \frac1{\sin{n \pi} \cos{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)} - \cos{n \pi} \sin{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)}}\\ &= \frac{(-1)^{n+1}}{\sin{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)}}\\ &= \frac{(-1)^{n+1}}{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)-\frac16 (n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)^3+\cdots}\end{align}$$

The residue is the coefficient of $\frac1{\zeta}$, which is $(-1)^{n+1}/(n^2 \pi^2)$. Within $|z|=1/5$, we only consider $|n| \ge 2$. Thus away from $z=0$ the contribution to the integral is

$$i 2 \pi (2) \sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n^2 \pi^2} = -i \frac{4}{\pi} \left (1-\frac{\pi^2}{12} \right)$$

There is also an essential singularity at $z=0$. $f$ may be written as

$$f(z) = \frac1{\frac1z - \frac16 \frac1{z^3} + \frac1{120} \frac1{z^5} - \cdots} $$

and thus has no residue here. Thus, the integral is

$$-i \left (\frac{4}{\pi} - \frac{\pi}{3} \right ) $$

(Thanks to @achillehui for pointing out a crucial mistake.)