Show that $\int_0 ^\infty e^{-x^2}\cos(2bx)\ \mathrm dx = \frac{\sqrt{\pi}e^{-b^2}}{2}$

Asked Oct 23 '22 at 19:10

Active Oct 23 '22 at 19:32

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My textbook says show that $$\int_0 ^\infty e^{-x^2}\cos(2bx)\ \mathrm dx = \frac{\sqrt{\pi}e^{-b^2}}{2}$$

Also, there is a hint. Integrate $f(z)=e^{-z^2}$ on rectangle given by $|x|\leq R,0\leq y \leq b$ and also use integral $\int_{-\infty} ^\infty e^{-t^2}dt=\sqrt\pi$

Also, I am reading this question answered by Moli.

And observing that on the two vertical sides, the integral approaches zero as $R→∞$, you will deduce that

$\int_{-\infty}^\infty{e^{-x^2}\cos(ax)dx}=\sqrt{\pi}\cdot e^{-\frac{a^2}{4}}.$

Can you please explain how this equality is derived? Why vertical integrals approach $0$ and how from that follows equality.

edited Oct 23 '22 at 19:32

Sumanta

9,534

asked Oct 23 '22 at 19:10

unit 1991

On this site I found something here and here and here and here, and more. – Matija Oct 23 '22 at 20:45
Did you actually try to follow the hint? What happens when you perform the contour integral over the boundary of the rectangle? – Mark Saving Oct 23 '22 at 20:53
Can you show us your attempt so we can see where exactly you're stuck? – Accelerator Oct 23 '22 at 21:09

Show that $\int_0 ^\infty e^{-x^2}\cos(2bx)\ \mathrm dx = \frac{\sqrt{\pi}e^{-b^2}}{2}$

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