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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.

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    $\begingroup$ Use \oint for the output $\oint$. See this. $\endgroup$
    – PinkyWay
    Commented May 17, 2020 at 21:29
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    $\begingroup$ Thank you! Always looking for new LaTeX tips. $\endgroup$
    – user711703
    Commented May 17, 2020 at 21:33
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    $\begingroup$ You're welcome! 😀 $\endgroup$
    – PinkyWay
    Commented May 17, 2020 at 21:34
  • $\begingroup$ 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 $z\in\left\{\pm\frac1\pi,\pm\frac1{2\pi},\pm\frac1{3\pi}\right\}$. $\endgroup$
    – robjohn
    Commented May 18, 2020 at 13:24

1 Answer 1

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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

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