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I’m trying to perform the contour integral $\oint f(z)\mathrm{d}z=\oint\frac1{(z^2+a^2)\sin\pi z}\mathrm{d}z$ where the contour comprises a straight line from $-\infty-\mathrm{i}a/2$ to $\infty-\mathrm{i}a/2$, as well as a semicircular contour in either the upper or lower half plane. I want to prove that the contribution from the semicircular contour goes to zero as I take the radius of the semicircle to go to infinity. My approach is to consider the estimation lemma as follows:

\begin{align} \left|\int_{\text{Semicircle}}\frac{\mathrm{d}z}{(z^2+a^2)\sin\pi z}\right|&\leq\sup_{z\in\text{Semicircle}}\pi R\left|\frac1{(z^2+a^2)\sin\pi z}\right|\\ &\sim\sup_{z\in\text{Semicircle}}R\cdot\mathcal{O}\left(\frac1{R^2}\right)|\csc\pi z|\\ &\sim\sup_{z\in\text{Semicircle}}\mathcal{O}\left(\frac1{R}\right)|\csc\pi z|. \end{align}

However, I don't see how $\csc z$ decays faster than $1/R$, or simplify it further to show that it decays. Is there a way to do so? Thanks!

user107224
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  • See here: https://math.stackexchange.com/q/1315970/269764 – Brevan Ellefsen May 25 '21 at 23:31
  • A box contour is natural because $\sin(\pi z)$ (where $z = x+iy$) decomposes as $\sin (\pi x)\cosh (\pi y)+i\cos(\pi x) \sinh (\pi y)$. Notice either $x$ or $y$ are fixed along the sides of the box, making the analysis rather easy. You could probably handle a semicircle contour similarly, though if I were to take that route I'd first try to rewrite it a bit and split into pieces to use Jordan's Lemma – Brevan Ellefsen May 25 '21 at 23:32

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