Here we consider a complex function $f(z) = \frac{\sin(z)}{z-\alpha}$ where $\alpha = \frac{\pi}{2}+i \log(2)$.
I want to know $\lim_{R\to\infty}\int_{-R}^{R}f(z)dz$ (let us call it $I$ for convenience).
Here is my solution so far:
(I got the residue of $f(z)$ in $z=\alpha$, which is $\frac{5}{4}$.)
Consider a positively oriented contour $\Gamma$ consisting of a line from $-R$ to $R$ (call it $\gamma_{-R,R}$) and an upper half circle with radius $R$ (call it $C_R^{+}$). Then, $$ \int_\Gamma f(z) dz = \int_{\gamma_{-R,R}}f(z)dz + \int_{C_R^{+}}f(z)dz. $$ By Cauchy's residue theorem, $$ \int_\Gamma f(z) dz = 2\pi i\cdot Res(\alpha) = \frac{5\pi i}{2}. $$ Hence, $$I = \frac{5\pi i}{2} - \lim_{R\to\infty}\int_{C_R^{+}}f(z)dz.$$
Here I stuck since I don't know how to find $\lim_{R\to\infty}\int_{C_R^{+}}f(z)dz$ which is integral over half circle (not closed).
Any helps or ideas (hints too) are more than welcome!