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Does this integral converge

$$\lim_{R \to \infty}\int_{\frac{\pi}{2}}^{\pi}\frac{e^{Rte^{i\theta}}}{\sqrt{Re^{i\theta}+1}}\cdot iRe^{i\theta}d\theta$$

where t is a positive integer?

If the integral diverges, how can I prove this?

p.s. This integral is part of a larger contour integral to calculate

$$\int_{a-i\infty}^{a+i\infty} \frac{e^{zt}}{\sqrt{1+z}} dz$$

I know that

$$\lim_{R \to \infty}\int_{\frac{\pi}{2}}^{\pi}\frac{e^{Rte^{i\theta}}}{\sqrt{Re^{i\theta}+1}}\cdot iRe^{i\theta}d\theta + \int_{-\pi}^{\frac{-\pi}{2}}\frac{e^{Rte^{i\theta}}}{\sqrt{Re^{i\theta}+1}}\cdot iRe^{i\theta}d\theta=0$$

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What is $R$? If $R\neq 1$, you are integrating over a finite domain of $\theta$ and there are no singularities. I feel like what you really want to ask is what happens to the integral as $R\rightarrow\infty$. – Alex R. Sep 13 '12 at 16:34
@Sam That is what I mean. I updated the question. – wnvl Sep 13 '12 at 16:36
up vote 4 down vote accepted


use $$\left| \frac{e^{Rt e^{i\theta}}}{\sqrt{R e^{i\theta} +1}} iR e^{i\theta} \right|\leq c\sqrt{R}\, e^{Rt \cos\theta}$$ with an appropriate $c$ for $R\geq R_0$ to show that the integral converges.

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