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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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    @Sam That is what I mean. I updated the question.2012-09-13

1 Answers 1

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Hint:

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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    @Zophikel: no Lemma of Jordan. The inequality simply follows from $|R/\sqrt{R e^{i\theta } + 1}|= \sqrt{R} /|\sqrt{1 +e^{-i\theta} /R}| \leq c \sqrt{R}$ if $R$ is large enough.2018-07-02