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Let $\gamma(z_0,R)$ denote the circular contour $z_0+Re^{it}$ for $0\leq t \leq 2\pi$. Evaluate $$\int_{\gamma(0,1)}\frac{\sin(z)}{z^4}dz.$$

I know that \begin{equation} \int_{\gamma(0,1)}\frac{\sin(z)}{z^4}dz = \frac{1}{z^4}\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots\right) = \frac{1}{z^3}-\frac{1}{6z}+\cdots \end{equation} but I'm not sure if I should calculate the residues and poles or to use Cauchy's formula?

Using Cauchy's formula would give $$ \frac{2\pi i}{1!} \frac{d}{dz}\sin(z),$$ evaluated at $0$ gives $2\pi i$? I'm not sure though, any help will be greatly appreciated.

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  • $\begingroup$ Since you divide by $z^4$, you need the third derivative in Cauchy's formula. So you'd get $$\frac{2\pi i}{3!} \left(\frac{d}{dz}\right)^3\Biggl\lvert_{z=0}\sin z.$$ $\endgroup$ Commented Jan 22, 2014 at 22:00
  • $\begingroup$ The residue is the coefficient of $z^{-1}$ in the Laurent series, $-{1\over 6}$ in this case. $\endgroup$ Commented Jan 22, 2014 at 22:32

2 Answers 2

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Cauchy's integral formula is

$$f^{(n)}(z) = \frac{n!}{2\pi i} \int_\gamma \frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta,$$

where $\gamma$ is a closed path winding once around $z$, and enclosing no singularity of $f$.

Thus in your example, $n = 3$, and you need the third derivative,

$$\int_{\gamma(0,1)} \frac{\sin z}{z^4}\,dz = \frac{2\pi i}{3!} \sin^{(3)} 0 = \frac{2\pi i}{6} (-\cos 0) = - \frac{\pi i}{3}.$$

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  • $\begingroup$ why is n=3? @Daniel Fischer $\endgroup$
    – R_Squared
    Commented Sep 9, 2023 at 12:07
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The function $f(z)=\frac{\sin{z}}{z^4}$ has a third order pole $z=0$ inside $\gamma$. By Cauchy's residue theorem, the integral $$ \int_{\gamma}f(z)\,dz=2{\pi}i\cdot\underset{z=0}{\rm res}\, f(z),$$ where the residue at the third order pole $z=0$ can be calculated using formula $$\underset{z=0}{\rm res}\, f(z)=\frac{1}{(3-1)!}\cdot\lim_{z\to 0}[z^3\cdot f(z)]^{(3-1)}=\frac{1}{2}\cdot\biggl(\frac{\sin{z}}{z}\biggr)''\biggr|_{z=0}=-\frac{1}{6}. $$ Alternatively, though in fact by defintition, the residiue at $z=0$ can be found as the $c_{-1}$ coefficient at the term $z^{-1}$ of the Laurent series in the ring $0<|z|<\infty$ for $f(z)$ having two isolated singularities on the complex plane: a pole at $z=0$ and an essential singularity at $z=\infty$. More precisely, $$f(z)=\frac{1}{z^4}\cdot (z-\frac{z^3}{3!}+\frac{z^5}{5!}-\dots)=\frac{1}{z^3}-\frac{1}{6z}+\frac{z}{5!}-\dots\,,\quad 0<|z|<\infty, $$ whence readily follows $$\underset{z=0}{\rm res}\, f(z)\overset{def}{=}c_{-1}=-\frac{1}{6}\,.$$ Hence, by Cauchy's residue theorem, the integral $$ \int_{\gamma}f(z)\,dz=-\frac{2{\pi}i}{6}=-\frac{{\pi}i}{3}\,.$$

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  • $\begingroup$ $\sin$ has a zero in $0$, the pole has order $3$. $\endgroup$ Commented Jan 22, 2014 at 23:54
  • $\begingroup$ Daniel Fischer, thank you for your kind remark. $\endgroup$
    – mkl314
    Commented Jan 23, 2014 at 0:25

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