Calculating an integral using complex analysis

Solution 1:

You don't need complex analysis to solve this. Note that we can split the integral as follows $$\int_0^\pi e^{-R\sin(t)} dt = \int_0^{\pi/2} e^{-R\sin(t)} dt + \int_{\pi/2}^\pi e^{-R\sin(t)} dt. $$

Since $\sin{t} > \frac{2}{\pi} t >0$ for $t\in (0,\pi/2)$ and $\sin{t} > -\frac{2}{\pi} t +2 >0$ for $t\in (\pi/2,\pi)$, we have $$\int_0^\pi e^{-R\sin(t)} dt < \int_0^{\pi/2} e^{-2Rt/\pi} dt + \int_{\pi/2}^\pi e^{-R(2-2t/\pi)} dt. $$

These integrals are easy to compute. One finds then that the integral can be bounded from above with functions in $R$ that are clearly going to zero for $R\to\infty$.This proves that your limit is indeed zero.