Is there a way to solve such and integral: $$\int_L\frac{\sin(\frac{1}{z})}{z(1-z)} \mathrm dz$$ $L$ can be:
- an arc of a semicircle with unit radius, centered at the origin and running clockwise from $1$ to $-1$ in the lower half of complex plane of variable $z$;
- an interval along the $\Re(z)$ axe from $1$ to $-1$.
By the way, I got used to the contours running counterclockwise. What do I need to change the direction? Change the limis of integration and invert the sign of the integral?
Ho to work with the essential singularity at $z=0$ in the econd case?
Changing variables to $\zeta=1/z$, this becomes $$\int_1^{-1} \frac{\sin\left(\frac1z\right)}{1-z}\frac{\text dz}{z} =\int_{-1}^{1} \frac{\sin\left(\zeta\right)}{\zeta-1}{\text d\zeta}.$$ This integral is divergent at $\zeta=1$ (so the original one was divergent at $z=1$). However, if you change the sine to $\sin\left(\frac\pi z\right)$ you introduce a zero at $\zeta=1$ which makes the integral convergent. Wolfram Alpha will happily calculate this for you. You can also reduce this, by changing to $t=1-\zeta$, to the canonical form of the sine integral, as $$\int_1^{-1} \frac{\sin\left(\frac\pi z\right)}{1-z}\frac{\text dz}{z} =\int_{-1}^{1} \frac{\sin\left(\pi\zeta\right)}{\zeta-1}{\text d\zeta} =-\int_{0}^{2} \frac{\sin\left(\pi t\right)}{t}{\text dt} =-\text{Si}(2\pi).$$ This cannot be reduced further in terms of elemental functions, but it is as easy to calculate, in most serious contexts, as any other special function.