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We know that the Gaussian integral is $$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}.$$ Now, we want to compute the following integral $$\int_{-\infty}^{\infty}e^{-at^2}\cos btdt=?$$ where $a>0$, $b$ is a real constant.

Could anybody give some helpful links or hints? Thx!

Update: $a>0$, sorry for the mistake!

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marked as duplicate by Fabian, Arkamis, muzzlator, Henry T. Horton, Amzoti Mar 21 '13 at 20:43

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With $a < 0$ this integral does not converge on $(-\infty, \infty)$. –  m0nhawk Mar 21 '13 at 12:11
I think you mean $\Re{a}>0$. –  Ron Gordon Mar 21 '13 at 12:13
Thanks for pointing out my mistake, $a$ should be bigger than 0. –  user39843 Mar 21 '13 at 12:18
See here also: math.stackexchange.com/questions/283610/… –  Byron Schmuland Mar 21 '13 at 12:22
Good, your link should be helpful. @ByronSchmuland –  user39843 Mar 21 '13 at 12:23

3 Answers 3

up vote 2 down vote accepted

$$\cos{b t} = \frac{1}{2} (e^{i b t} + e^{-i b t})$$

Complete the square in the exponential. For example,

$$a t^2 + i b t = a \left ( t^2 + i \frac{b}{a} t -\frac{b^2}{4 a^2}\right ) + \frac{b^2}{4 a}$$

Then use the fact that

$$\int_{-\infty}^{\infty} dx \: e^{-a x^2} = \sqrt{\frac{\pi}{a}}$$

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I don't think $cosbt=1/2(e^{ibt}+e^{-ibt})$ it's $coshbt$ –  Mr.ØØ7 Mar 21 '13 at 12:16
@exploringnet: uhhh....not so much. –  Ron Gordon Mar 21 '13 at 12:17
It would work for $\cos bt=1/2(e^ibt +e^{-ibt})$, wouldn't it? @exploringnet –  user39843 Mar 21 '13 at 12:25
I don't know.. :P We have not studied this relation. –  Mr.ØØ7 Mar 21 '13 at 12:28
math.stackexchange.com/questions/188257/… your link gives a elegant solution of $\int_0^\infty e^{- \left ( x - \frac a x \right )^2} dx$, but I can not get the point that how to evaluate mine? Their relationship seems not so evident. –  user39843 Mar 21 '13 at 12:38

Let us assume that $a=1$. Show by using Fubini that $$ \int_0^\infty \sin(bt)\cdot te^{-t^2/2}\,\mathrm dt=b\int_0^\infty\cos(bt)\cdot e^{-t^2/2}\,\mathrm dt $$ for all $b\in\mathbb{R}$. Use this to deduce that $$ p:b\mapsto \int_{\mathbb{R}}\cos(bt)\cdot\frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,\mathrm dt $$ solves the differential equation $$ p'(b)=-b\cdot p(b),\quad p(0)=1 $$ or in other words $p(b)=e^{-b^2/2}$.

For a general $a>0$ you could use the same argument.

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Wow! Elegant proof! Good! @Stefan Hansen –  user39843 Mar 21 '13 at 13:32

Try to find below book. It has answers for a lot of integrals. I think this will be there too.


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Well, I just need to know how to compute this guy. Thx! –  user39843 Mar 21 '13 at 12:20
@Norman This is not a answer(nearly -1). Not even a good comment. –  user45099 Mar 21 '13 at 13:00

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