How to calculate $\int_{C(0,1)}\frac{\sin z}{z^4}dz$

Question

$\displaystyle\int_{C(0,1)}\frac{\sin z}{z^4}\:\mathrm{d}z$, where $C(0,1)$ is the circle around $0$ with radius $1$

$$\displaystyle\int_{C(0,1)}\frac{\sin z}{z^4}\:\mathrm{d}z=\int_0^{2\pi}i\frac{\sin(e^{it})}{e^{3it}}\:\mathrm{d}t$$ seems difficult to solve. Do I have to use some trigonometric identities and hyperbolic functions?

Answer 1

$$\int_{C(0,1)}\frac{\sin z}{z^4}\:\mathrm{d}z\\ =\int_{C(0,1)}\frac{z- \frac16z^3+\frac1{120}z^5-\dots}{z^4}\:\mathrm{d}z\\ =\int_{C(0,1)}\left(\frac{1}{z^3}-\frac1{6z}+\frac1{120}z-\dots\right)\:\mathrm{d}z $$ Only the contour containing the $-\frac1{6z}$ is non-zero. Hence $$\int_{C(0,1)}\frac{\sin z}{z^4}\:\mathrm{d}z\\ =-\frac16\int_{C(0,1)}\frac1{z}\:\mathrm{d}z\\ =-\frac13i\pi$$

Answer 2

I think this is using the Cauchy integral formula

\begin{equation*} f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta \end{equation*}

where $\gamma$ must contain no singularity of $z.$ Here, $z=0$ is a pole of order $4.$

Here, $n=3$ so we need the third derivative:

\begin{equation*} \int_{C(0,1)}\frac{\sin(z)}{z^4}dz=\frac{2\pi i}{3!}(-\cos(0))=-\frac{\pi i}{3}. \end{equation*}

I think this is the way to approach this problem. Is this okay?