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I try show, that the following integral (for $0<a<1$) gives:

$\int_0^\infty e^{-(1-a^2)x^2} \cos(2ax^2)dx = \frac{\sqrt{\pi}}{2(1+a^2)}$

I guess the way to go is complex Integration and using the fact, that this is the same as $\frac{1}{2}\int_{- \infty}^\infty e^{-(1-a^2)x^2} \cos(2ax^2)dx $

First I tried to use a rectangular shaped path in the complex plane, which I sucessfully used at a similar problem a while ago. Poorly I didn't remember, that this only seems to work for a linear term $2ax$ i the cosine.

EDIT: This way is shown in a very similar way as I did in: Evaluatig: $\int_{0}^{\infty}{e^{ax^2}\cos(bx)dx}$ But as mentioned - this doesn't seem to work in my case...

My second idea was to get some hints for the path I should use via concerning the given result. So I guess (because there is $(1+a^2)$ in the denominator) I need to integrate $e^{-(1+a^2)^2x^2}$ to get this term via using the Gaussian integral $\int_{-\infty}^{\infty} e^{-bx^2} dx = \sqrt{\frac{\pi}{b}}$. Although I tried to find some complete square I wasn't able to produce the necessary terms (without getting loads of waste, which makes it difficult again)

I would be very grateful, if anyone could help me or give me a hint how to solve this integral!

Thanks!

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    $\begingroup$ I would try using the exponential form of the cosine and split the integral into two. $\endgroup$ – aleden Apr 22 '18 at 21:38
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$$ \begin{eqnarray*} \frac{1}{2} \int_{-\infty}^\infty dx \; e^{-(1-a^2)x^2} \cos 2ax^2 &=& \frac{1}{4} \int_{-\infty}^\infty dx \; e^{-(1-a^2)x^2} (e^{i 2ax^2} + e^{-i 2ax^2})\\ &=& \frac{1}{4} \int_{-\infty}^\infty dx \; e^{-(1-a^2-2ia)x^2} + e^{-(1-a^2+2ia)x^2}\\ &=& \frac{1}{4} \int_{-\infty}^\infty dx \; e^{-(1-ia)^2 x^2} + e^{-(1+ia)^2 x^2}\\ &=& \frac{1}{4} ( \frac{\sqrt{\pi}}{1-ia} + \frac{\sqrt{\pi}}{1+ia})\\ &=& \frac{\sqrt{\pi}}{4} \frac{2}{1+a^2} \end{eqnarray*} $$

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  • $\begingroup$ Note you could've gotten away with just taking the real part of one complex integral or the other instead of having to sum them. $\endgroup$ – Ian Apr 22 '18 at 22:51
  • $\begingroup$ (+1) It took me a while to post my answer, but I see that my answer is very similar to yours. I will leave mine as it is slightly different and provides a bit more explanation. $\endgroup$ – robjohn Apr 22 '18 at 23:25
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$$\newcommand{\Re}{\operatorname{Re}} \begin{align} \int_0^\infty e^{-\left(1-a^2\right)x^2}\cos\left(2ax^2\right)\,\mathrm{d}x &=\frac12\int_0^\infty e^{-\left(1-a^2\right)x^2}\left(e^{i2ax^2}+e^{-i2ax^2}\right)\mathrm{d}x\tag1\\ &=\frac12\int_0^\infty\left(e^{-[(1-ia)x]^2}+e^{-[(1+ia)x]^2}\right)\mathrm{d}x\tag2\\ &=\Re\left(\int_0^\infty e^{-[(1+ia)x]^2}\mathrm{d}x\right)\tag3\\ &=\Re\left(\frac1{1+ia}\int_0^{(1+ia)\infty}e^{-x^2}\mathrm{d}x\right)\tag4\\ &=\Re\left(\frac1{1+ia}\int_0^\infty e^{-x^2}\mathrm{d}x\right)\tag5\\ &=\frac{\sqrt\pi/2}{1+a^2}\tag6 \end{align} $$ Explanation:
$(1)$: $\cos(x)=\frac12\left(e^{ix}+e^{-ix}\right)$
$(2)$: combine and factor the exponents
$(3)$: $\Re(z)=\frac12\left(z+\bar{z}\right)$
$(4)$: substitute $x\mapsto\frac{x}{1+ia}$
$(5)$: The integral along $(1+ia)[0,R]$ equals the integral along
$\phantom{(5)\text{:}}$ $[0,R]\cup [R,R(1+ia)]$ and the integral along $[R,R(1+ia)]$
$\phantom{(5)\text{:}}$ vanishes as $R\to\infty$
$(6)$: $\Re\left(\frac1{1+ia}\right)=\frac1{1+a^2}$

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