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I have the following integral $$\int_{-\infty}^{\infty}e^{-x^2}\cos(2bx)\,\mathrm{d}x$$ How can I prove this integral equals $\sqrt{\pi}e^{-b^2}$ by using the residue theorem?

For $b>0$ and $\gamma_R$ a middle circle of radius $R$ up $x$-axis, I tried doing \begin{align*} \oint_{\gamma_R}e^{-z^2}e^{2ibz}\,\mathrm{d}z&=0 \end{align*} since $f(z)=e^{-z^2}e^{2ibz}$ is continous over $\gamma_R$.

Diego A. Alvarado Silos
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    Check this: https://math.stackexchange.com/a/1363059/42969. – Martin R Oct 10 '20 at 20:51
  • Next you need to investigate whether the part on the semicircle goes to $0$ as $R \to \infty$. This will tell us whether this contour is a good one to use. – GEdgar Oct 10 '20 at 20:52
  • After completing the square and a change of variables $z=x-ib$ you basically want to prove $$\int_{-\infty}^\infty e^{-z^2} , {\rm d}z = \sqrt{\pi} , .$$ You can shift the contour as you like, since the integrand is holomorphic. But other than that what makes you think contour integration is a good approach to calculate the latter integral? – Diger Oct 10 '20 at 21:24

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