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In calculating with the residue theorem I find $$\int_{-\infty}^\infty \frac{e^{-z^2}}{1+z^2} \, dx = \pi e$$

Online, a video calculating this integrand by means of "Feynman's trick" produces another term, $-\pi \cdot e\cdot \mathrm{erf}(1)$. I find only one term from the Residue at $(0,i)$ in the upper half-plane. The limit of the integral on the semicircular arc, I find goes $0$ as the radius goes to infinity. Have I neglected something?

PrincessEev
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goedelite
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    Why does the integral over the semicircle go to zero? For example if we fix the real part of $z$ then $e^{-z^2}/(1+z^2)$ is unbounded as $\operatorname{im} z\to+\infty$. – leoli1 Jan 15 '21 at 20:08
  • Because I am a dummy! :) – goedelite Jan 15 '21 at 20:13

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