Integral $\int_{\sigma}{{\sin(z)dz}\over{z^3}}$ with $\sigma:[0,2\pi] \to C, \sigma (t)=e^{it}$

Question

How would one solve it without residues? I tried using the definition of integral in a curve but it lead me nowhere. What else could I try?

Answer 1

Note that without appeal to residues, we can write

$$\begin{align} \oint_\sigma \frac{\sin(z)}{z^3}\,dz&=i\int_0^{2\pi}\sin(e^{it}) e^{-i2t}\,dt\\\\ &=i\int_0^{2\pi} \sum_{n=1}^\infty \frac{(-1)^{n-1}e^{i(2n-1)t}}{(2n-1)!}\,e^{-i2t}\,dt \tag 1\\\\ &=i\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)!}\color{blue}{\underbrace{\int_0^{2\pi} e^{i(2n-3)t}\,dt}_{=0\,\,\text{for all}\,\,n}} \tag 2\\\\ &=0 \end{align}$$

where the uniform convergence of the series justifies interchanging the integration and summation operators in going from $(1)$ to $(2)$.