I need proof .$\int^{\infty}_{0} e^{-at^2} \cos(2 x t) \, dt = \frac{1}{2} \sqrt{\frac{\pi}{a}} e^{\frac{-x^2}{a}}$. $$\int^{\infty}_{0} e^{-at^2}  \cos(2 x t) \, dt = \frac{1}{2} \sqrt{\frac{\pi}{a}} e^{\frac{-x^2}{a}}$$
for Re$(a)>0$,  for all real value greater than zero.
I took this equation from the book, Abramowitz and Stegun, Equation number $7.4.6$
 A: Hint
Consider that you look for the real part of
$$\int^{\infty}_{0} e^{-at^2} e^{2i x t} \, dt=\int^{\infty}_{0}e^{-a t^2+2 i t x}\,dt$$
Complete the square to face a well known integral.
A: Well, solving a more general problem we are looking at:
$$\mathcal{I}_\text{n}\left(\alpha\right):=\int_0^\infty\exp\left(-\alpha\cdot x^2\right)\cdot\cos\left(\text{n}\cdot x\right)\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can rewrite:
$$\mathcal{I}_\text{n}\left(\alpha\right)=\int_0^\infty\mathcal{L}_x\left[\cos\left(\text{n}\cdot x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\exp\left(-\alpha\cdot x^2\right)\right]_{\left(\text{s}\right)}\space\text{d}\text{s}\tag2$$
Using the table of selected Laplace transforms, we get:
$$\mathcal{I}_\text{n}\left(\alpha\right)=\int_0^\infty\frac{\text{s}}{\text{n}^2+\text{s}^2}\cdot\mathcal{L}_x^{-1}\left[\exp\left(-\alpha\cdot x^2\right)\right]_{\left(\text{s}\right)}\space\text{d}\text{s}\tag3$$
Using:
$$\exp\left(x\right):=e^x=\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\tag4$$
We can write:
$$\mathcal{I}_\text{n}\left(\alpha\right)=\int_0^\infty\frac{\text{s}}{\text{n}^2+\text{s}^2}\cdot\mathcal{L}_x^{-1}\left[\sum_{\text{k}=0}^\infty\frac{\left(-\alpha\cdot x^2\right)^\text{k}}{\text{k}!}\right]_{\left(\text{s}\right)}\space\text{d}\text{s}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\alpha^\text{k}}{\text{k}!}\cdot\int_0^\infty\frac{\text{s}}{\text{n}^2+\text{s}^2}\cdot\mathcal{L}_x^{-1}\left[x^{2\text{k}}\right]_{\left(\text{s}\right)}\space\text{d}\text{s}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\alpha^\text{k}}{\text{k}!}\cdot\int_0^\infty\frac{\text{s}}{\text{n}^2+\text{s}^2}\cdot\frac{1}{\text{s}^{1+2\text{k}}}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\space\text{d}\text{s}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\alpha^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\int_0^\infty\frac{1}{\text{n}^2+\text{s}^2}\cdot\frac{1}{\text{s}^{2\text{k}}}\space\text{d}\text{s}\tag5$$

now, use this answer to prove the final answer.

A: start with
$$I\left( x \right)=\int_{0}^{\infty }{{{e}^{-a{{t}^{2}}}}}\cos \left( 2xt \right)dt$$
then differentiation under the integral
$${I}'\left( x \right)=\int_{0}^{\infty }{\frac{\partial }{\partial x}\left[ {{e}^{-a{{t}^{2}}}}\cos \left( 2xt \right) \right]dt}=-\int_{0}^{\infty }{2t{{e}^{-a{{t}^{2}}}}}\sin \left( 2xt \right)dt$$
integration by parts
$${I}'\left( x \right)=-\int_{0}^{\infty }{\overbrace{2t{{e}^{-a{{t}^{2}}}}}^{dv}}\overbrace{\sin \left( 2xt \right)}^{u}dt=-\left. \sin \left( 2xt \right)\frac{{{e}^{-a{{t}^{2}}}}}{-a} \right|_{0}^{\infty }+\int_{0}^{\infty }{\frac{{{e}^{-a{{t}^{2}}}}}{-a}2x\cos \left( 2xt \right)dt}$$
the first term vanishes and we have a simple differential equation
$${I}'\left( x \right)=-\frac{2x}{a}I\left( x \right)$$
with solution
$$I\left( x \right)=I\left( 0 \right){{e}^{\frac{-{{x}^{2}}}{a}}}$$
A: $$
\begin{align}
\int_0^\infty e^{-at^2}\cos(2xt)\,\mathrm{d}t
&=\frac12\int_{-\infty}^\infty e^{-at^2}\cos(2xt)\,\mathrm{d}t\tag1\\
&=\frac12\int_{-\infty}^\infty e^{-at^2}e^{2ixt}\,\mathrm{d}t\tag2\\
&=\frac12e^{-x^2/a}\int_{-\infty}^\infty e^{-a\left(t-ix/a\right)^2}\,\mathrm{d}t\tag3\\
&=\frac12\sqrt{\frac\pi{a}}e^{-x^2/a}\tag4
\end{align}
$$
$(1)$: symmetry
$(2)$: $\sin(2xt)$ is odd in $t$
$(3)$: complete the square
$(4)$: Cauchy's Integral Theorem and $\int_{-\infty}^\infty e^{-at^2}\,\mathrm{d}t=\sqrt{\frac\pi{a}}$ 
