Compute $\oint_{C_{o}} \frac{1}{\sin \left(\frac{1}{z}\right)} d z-\oint_{C_{i}} \frac{1}{\sin \left(\frac{1}{z}\right)} \mathrm d z$

Compute the integral $$\oint_{C_{o}} \frac{1}{\sin \left(\frac{1}{z}\right)} d z-\oint_{C_{i}} \frac{1}{\sin \left(\frac{1}{z}\right)} \mathrm d z$$ where the outer circle, $$C_{o},$$ is the unit circle and the inner circle, $$C_{i},$$ is the circle of radius $$1 / 10$$ centered at the origin. (HINT: What region is enclosed by these two curves?)

I've adapted the power series of the given function from the Maclaurin series for the sine:

$$\left(\frac{1}{x} - \frac{1}{x^3} \frac{1}{3!} + \frac{1}{x^5} \frac{1}{5!} - \right)^{-1}$$.

If I were to sketch this region I would find that this is the annular region less than radius R=1 and greater than radius r=$$\frac{1}{10}$$, making the region $$\frac{1}{10}<|z|<1$$. I believe the outer circle would be the Maclaurin series of the function given $$|z|<1$$ minus the inner circle which would be a Laurent series since $$|z|>\frac{1}{10}$$. Are there any flaws in my thinking of which I should be aware? I feel as if I'm missing a small piece in order to correctly arrange these steps.

Clarification: I'm aware I take the residues of each and multiply them by $$2\pi i$$. I omitted these steps for clarity.

• Use \oint for the output $\oint$. See this. Commented May 17, 2020 at 21:29
• Thank you! Always looking for new LaTeX tips.
– user711703
Commented May 17, 2020 at 21:33
• You're welcome! ðŸ˜€ Commented May 17, 2020 at 21:34
• While the series $\left(\frac1z-\frac1{z^3}\frac1{3!}+\frac1{z^5}\frac1{5!}-\dots\right)^{-1}$ will converge for all $z\ne0$, it does little to help compute the residues at the singularities inside the annulus $\frac1{10}\le|z|\le1$, which are at $z\in\left\{\pm\frac1\pi,\pm\frac1{2\pi},\pm\frac1{3\pi}\right\}$.
– robjohn
Commented May 18, 2020 at 13:24

Evaluation of the Integral

Since each singularity (except the essential singularity at $$z=0$$, which is not contained in the combined contour) is simple, $$\newcommand{\Res}{\operatorname*{Res}} \Res_{z\in\left\{\pm\frac1\pi,\pm\frac1{2\pi},\pm\frac1{3\pi}\right\}}\frac1{\sin\left(\frac1z\right)}=-\frac{z^2}{\cos\left(\frac1z\right)}$$ The sum of the residues in the annulus is $$\frac2{\pi^2}-\frac2{4\pi^2}+\frac2{9\pi^2}=\frac{31}{18\pi^2}$$ Assuming the paths circle counterclockwise, the integral is $$2\pi i$$ times the sum of the residues $$\bbox[5px,border:2px solid #C0A000]{\frac{31i}{9\pi}}$$

Concerning Residues at Simple Singularities

Suppose that $$f(z)$$ has a simple zero at $$z=\alpha$$, then $$f'(\alpha)\ne0$$ and $$f(z)=(z-\alpha)f'(\alpha)+O(z-\alpha)^2$$ so that $$\frac1{f(z)}=\frac1{(z-\alpha)f'(\alpha)}+O(1)$$ Therefore, $$\Res_{z=\alpha}\frac1{f(z)}=\frac1{f'(\alpha)}$$

Mathematica Verification

z[r_,t_] := r Exp[I t]; Subtract @@ (NIntegrate[ 1/Sin[1/z[#,t]] I z[#,t], {t,0,2Pi}, WorkingPrecision->20]& /@ {1,1/10})

yields 6.*10^-21 + 1.0964007190775012020 I

N[31I/(9Pi),20] yields 1.0964007190775012020 I