Compute \[ \frac1{2\pi i}\int_C\frac{dz}{\sin(1/z)}, \] where \(C\) is the positively oriented circle \(|z|=1/5\).
Solution
1. Under the change of variables\[ w=\frac1z, \]the contour \(C\) becomes the circle\[ \abs w=5 \]with negative orientation.
Parametrize \[ z(t)=\frac15e^{it}, \qquad 0\leq t\leq2\pi. \] Then \[ w(t)=\frac1{z(t)}=5e^{-it}, \] which traverses \(\abs w=5\) clockwise.
2. The original contour integral satisfies\[ \int_C\frac{dz}{\sin(1/z)} = \int_{\abs w=5}^{+} \frac{dw}{w^2\sin w}, \]where the circle on the right is positively oriented.
Since \[ z=\frac1w, \qquad dz=-\frac{dw}{w^2}, \] step 1 gives \[ \int_C\frac{dz}{\sin(1/z)} = \int_{\abs w=5}^{-} -\frac{dw}{w^2\sin w}. \] Reversing the orientation removes the minus sign: \[ \int_{\abs w=5}^{-} -\frac{dw}{w^2\sin w} = \int_{\abs w=5}^{+} \frac{dw}{w^2\sin w}. \]
3. The meromorphic function\[ F(w)=\frac1{w^2\sin w} \]has exactly three poles inside \(\abs w<5\), at\[ 0,\quad \pi,\quad -\pi. \]
Away from \(w=0\), the poles occur at the zeros of \(\sin w\), namely \[ w=k\pi, \qquad k\in\ZZ. \] Inside \(\abs w<5\), the only nonzero such points are \(\pi\) and \(-\pi\), because \[ \pi<5<2\pi. \] The factor \(w^2\) also makes \(w=0\) a pole.
4. The residue of \(F\) at \(w=\pi\) is\[ -\frac1{\pi^2}. \]
The zero of \(\sin w\) at \(\pi\) is simple, with \[ \cos\pi=-1. \] Therefore \[ \begin{aligned} \operatorname{Res}_{w=\pi}F(w) &= \frac1{\pi^2} \frac1{\cos\pi}\\ &= -\frac1{\pi^2}. \end{aligned} \]
5. The residue of \(F\) at \(w=-\pi\) is\[ -\frac1{\pi^2}. \]
Again the zero of \(\sin w\) is simple, and \[ \cos(-\pi)=-1. \] Thus \[ \operatorname{Res}_{w=-\pi}F(w) = \frac1{(-\pi)^2} \frac1{\cos(-\pi)} = -\frac1{\pi^2}. \]
6. The residue of \(F\) at \(w=0\) is\[ \frac16. \]
The Taylor expansion \[ \sin w = w-\frac{w^3}{6}+O(w^5) \] gives \[ \begin{aligned} \frac1{w^2\sin w} &= \frac1{ w^3 \left( 1-\frac{w^2}{6}+O(w^4) \right) }\\ &= \frac1{w^3} \left( 1+\frac{w^2}{6}+O(w^4) \right)\\ &= \frac1{w^3} + \frac1{6w} + O(w). \end{aligned} \] The coefficient of \(w^{-1}\) is \(1/6\).
7. The sum of the residues inside \(\abs w=5\) is\[ \frac16-\frac2{\pi^2}. \]
Add the residues from steps <1>4–<1>6: \[ \frac16-\frac1{\pi^2}-\frac1{\pi^2} = \frac16-\frac2{\pi^2}. \]
8. The value of the requested integral is\[ \boxed{ \frac1{2\pi i} \int_C \frac{dz}{\sin(1/z)} = \frac16-\frac2{\pi^2}. } \]
By steps <1>2–<1>3, the residue theorem applies to \(F\) on the positively oriented circle \(\abs w=5\). Hence \[ \frac1{2\pi i} \int_C \frac{dz}{\sin(1/z)} = \sum_{\abs a<5}\operatorname{Res}_{w=a}F(w). \] Step 7 evaluates this sum.
9. Q.E.D.
Step 8 is the requested value.