How to prove that $\lim_{R\rightarrow\infty}\int_{0}^{\pi}\frac{e^{iaRe^{i\theta}}iRe^{i\theta}d\theta}{b^2+R^2e^{2i\theta}}=0$ (Not sure how this will render in the title, so here it is again)
I'm looking to show that
$$
\lim_{R\rightarrow\infty}\int_{0}^{\pi}\frac{e^{iaRe^{i\theta}}iRe^{i\theta}d\theta}{b^2+R^2e^{2i\theta}}=0
$$
but I'm not quite sure how to proceed. Repeated application of Euler's formula doesn't seem to help, as you encounter a lot of complex infinities.
 A: We have $iaR e^{i \theta} = iaR \cos(\theta) - aR \sin(\theta)$. Hence, $\left\vert e^{iaR e^{i \theta}} \right \vert = e^{-aR \sin(\theta)}$. 
Also, $$\left\vert iRe^{i \theta} \right\vert = R$$ and $$\left\vert b^2 + R^2 e^{2i\theta} \right\vert > R^2 - b^2$$
Put these all together to get what you want.
A: Hint: Show that the absolute value of the integrand gets arbitrarily small as $R$ grows. This is easier once you get rid of things that vanish on taking the absolute value: $$|\text{integrand}| = \left|\frac{e^{iaRe^{i\theta}}iRe^{i\theta}}{b^2+R^2e^{2i\theta}}\right| = \left|\frac{e^{-aR\text{sin}(\theta)}R}{b^2+R^2e^{2i\theta}}\right|$$ 
Show that for large enough $R$ this is at most $\varepsilon$; then the absolute value of the integral will be at most $\pi \varepsilon$; i.e., it goes to zero as $R$ grows.
A: Note that the magnitude of the integral is bounded by
$$\frac{2 R}{R^2-b^2} \int_0^{\pi/2} d\theta \, e^{-a R \sin{\theta}} \le \frac{2 R}{R^2-b^2} \int_0^{\pi/2} d\theta \, e^{-2 a R \theta/\pi} \le \frac{\pi}{a (R^2-b^2)}$$
which vanishes as $\pi/(a R^2)$ as $R \to\infty$ for $a>0$.  (I used the inequality $\sin{\theta} \ge 2\theta/\pi$ for $\theta \in [0,\pi/2]$.)
