Given the straight line in the complex plane: $b+iR$ to $b+1+iR$ where $0<b<1$ and $|Im(a)|<\pi$, show the following:
$$\int_{C} \frac{e^{az}}{\sin(\pi z)} \,dz \rightarrow 0 \quad as \quad R \rightarrow \infty$$
I proceed by bounding the integral with $z=b+x+iR \quad ; \quad 0\leq x \leq 1$
$$\bigg|\int_{C} \frac{e^{az}}{\sin(\pi z)} \,dz\bigg|=\int_{b}^{b+1} \frac{|e^{a(b+x+iR)}|}{|\sin(\pi (b+x+iR))|} \,|dx| \leq \frac{2e^{ab}}{e^{-R}-e^{R}}$$
However, the last expression derived does not approach $0$, and I am unsure where I am going wrong -advice?
Edit: Just realized my process was indeed correct, and that the expression does tend to $0$.