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Let's say I'm given an integral $$\int_C \frac{z \text{d}z}{(z + 1)^3(z + i)}$$ over $C \equiv \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a, b > 0, a, b \neq 1$ in Gaussian plane.

My straight approach would be to use Residue theorem on poles $z = -1$ and $z = -i$ in dependence on the values of $a$ and $b$.

But recently we were doing an integral of $\frac{\sin x}{x}$ in our complex analysis class and we were using the half-circle trick with a small half circle around point $(-\epsilon, \epsilon)$.

Now, I've got confused about that trick. Do I understand it correctly that it was done because we were going through the real line with our contour, where we have a singularity and in the case of above integral I can just directly use Residue theorem on poles because my ellipse itself is not going through them?

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    $\begingroup$ More or less. $\frac{\sin z}{z}$ is entire, so there's no singularity, but to get a situation where the integral over the large semicircle tends to $0$ as the radius tends to $\infty$ one looks at $\frac{e^{iz}}{z}$ (or at $\frac{e^{-iz}}{z}$), and that has a singularity on the contour. Here we have no singularity on the closed contour and the residue theorem can be applied directly. $\endgroup$ – Daniel Fischer Jun 9 '18 at 13:08

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