Proving $\int_0^\infty e^{-(1-a^2)x^2} \cos(2ax^2)dx = \frac{\sqrt{\pi}}{2(1+a^2)}$ 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!
 A: $$
\begin{eqnarray*}
\frac{1}{2} \int_{-\infty}^\infty dx \; e^{-(1-a^2)x^2} \cos 2ax^2 &=& \frac{1}{4} \int_{-\infty}^\infty dx \; e^{-(1-a^2)x^2} (e^{i 2ax^2} + e^{-i 2ax^2})\\
&=& \frac{1}{4} \int_{-\infty}^\infty dx \; e^{-(1-a^2-2ia)x^2} + e^{-(1-a^2+2ia)x^2}\\
&=& \frac{1}{4} \int_{-\infty}^\infty dx \; e^{-(1-ia)^2 x^2} + e^{-(1+ia)^2 x^2}\\
&=& \frac{1}{4} ( \frac{\sqrt{\pi}}{1-ia} + \frac{\sqrt{\pi}}{1+ia})\\
&=& \frac{\sqrt{\pi}}{4} \frac{2}{1+a^2}
\end{eqnarray*}
$$
A: $$\newcommand{\Re}{\operatorname{Re}}
\begin{align}
\int_0^\infty e^{-\left(1-a^2\right)x^2}\cos\left(2ax^2\right)\,\mathrm{d}x
&=\frac12\int_0^\infty e^{-\left(1-a^2\right)x^2}\left(e^{i2ax^2}+e^{-i2ax^2}\right)\mathrm{d}x\tag1\\
&=\frac12\int_0^\infty\left(e^{-[(1-ia)x]^2}+e^{-[(1+ia)x]^2}\right)\mathrm{d}x\tag2\\
&=\Re\left(\int_0^\infty e^{-[(1+ia)x]^2}\mathrm{d}x\right)\tag3\\
&=\Re\left(\frac1{1+ia}\int_0^{(1+ia)\infty}e^{-x^2}\mathrm{d}x\right)\tag4\\
&=\Re\left(\frac1{1+ia}\int_0^\infty e^{-x^2}\mathrm{d}x\right)\tag5\\
&=\frac{\sqrt\pi/2}{1+a^2}\tag6
\end{align}
$$
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 $R\to\infty$
$(6)$: $\Re\left(\frac1{1+ia}\right)=\frac1{1+a^2}$
