Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$ Prove that :
$$ \frac{\sqrt{\pi}}{2} e^{-\frac{a^2}{4} } =\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$$

the only thing I can think of is differentiating the RHS and trying to get :
$$ -2 f'(a) =a f(a) $$
But I couldn't do it. 
Can anyone show me how to do this ? 
 A: One approach is to integrate $e^{-z^2}$ around a rectangle in the complex plane with vertices at $R,R+ia,-R+ia$ and $-R$.
Another approach is to expand $\cos ax$ in a Maclaurin series and then switch the order of integration and summation. You'll end up with a constant times the Maclaurin series of $e^{\frac{-a^{2}}{4}}$.
A: A nice simple, but useful, modification of this is that:
$$
\int_{-\infty}^{\infty} \frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sqrt{2\pi}\sigma}\cos(ax)dx = \exp\left(-\frac{\sigma^2 a^2}{2}\right)
$$
which helps for developing the Fourier series of the Gaussian.
A: Differential Equation Approach
Note that since the integrand is even we have
$$
\int_0^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x
=\frac12\int_{-\infty}^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x\tag1
$$
Then differentiating with respect to $a$ yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}a}\int_{-\infty}^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x
&=-\int_{-\infty}^\infty e^{-x^2}x\sin(ax)\,\mathrm{d}x\tag{2a}\\
&=\frac12\int_{-\infty}^\infty\sin(ax)\,\mathrm{d}e^{-x^2}\tag{2b}\\
&=-\frac12\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}\sin(ax)\tag{2c}\\
&=-\frac a2\int_{-\infty}^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x\tag{2d}\\
\end{align}
$$
$\text{(2a)}$: differentiate
$\text{(2b)}$: prepare to integrate by parts
$\text{(2c)}$: integrate by parts
$\text{(2d)}$: $\mathrm{d}\sin(ax)=a\cos(ax)\,\mathrm{d}x$
Since $f'(a)=-\frac a2f(a)$ has the solution $f(a)=ce^{-a^2/4}$ and $f(0)=\int_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}2$,
$$
\int_0^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x=\frac{\sqrt\pi}{2}\,e^{-a^2/4}\tag3
$$

Contour Integral Approach
$$
\begin{align}
\int_0^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x\tag{4a}\\
&=\frac12\int_{-\infty}^\infty e^{-x^2}e^{iax}\,\mathrm{d}x\tag{4b}\\
&=\frac12e^{-a^2/4}\int_{-\infty}^\infty e^{-\left(x-ia/2\right)^2}\,\mathrm{d}t\tag{4c}\\[3pt]
&=\frac{\sqrt\pi}2\,e^{-a^2/4}\tag{4d}
\end{align}
$$
$\text{(4a)}$: symmetry
$\text{(4b)}$: $\sin(ax)$ is odd in $x$
$\text{(4c)}$: complete the square
$\text{(4d)}$: Cauchy's Integral Theorem and $\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}t=\sqrt\pi$
A: You can integrate
$$f(z)=e^{-z^2},z\in\mathbb{C}$$
On the boundary of the rectangle
$$\{z\in\mathbb{C}:\Re(z)\in[-R,R]\wedge \Im(z)\in[0,a/2]\}$$
Given $a>0,R>a/2.$
Being $f$ entire, this integral is equal to zero.
Using the fact that
$$\int_0^\infty{e^{-x^2}dx}=\frac{\sqrt\pi}{2}$$
And observing that on the two vertical sides the integral approaches zero as $R \to \infty$, you will deduce that
$$\int_{-\infty}^\infty{e^{-x^2}\cos(ax)dx}=\sqrt{\pi}\cdot e^{-\frac{a^2}{4}},$$
And also
$$\int_{-\infty}^\infty{e^{-x^2}\sin(ax)dx}=0.$$
Therefore, by the fact that the function is even, we have that
$$\int_{0}^\infty{e^{-x^2}\cos(ax)dx}=\frac{\sqrt{\pi}}{2}\cdot e^{-\frac{a^2}{4}}.$$
A: Instead of cosine, take $e^{iax}$ and complete the square. This thing says that the Fourier transform of the standard normal distribution is itself.
A: Notice that
$$
  a f(a) + 2 f^\prime(a) = \int_0^\infty \exp(-x^2) \left( a \cos(a x) - 2 x \sin(a x) \right) \mathrm{d} x = \int_0^\infty \frac{ \mathrm{d}}{\mathrm{d}x} \left( \mathrm{e}^{-x^2} \sin\left(a x\right) \right) \mathrm{d} x = 0
$$
It now only remains to find $f(0) = \int_0^\infty \exp(-x^2) \mathrm{d}x = \frac{1}{2} \int_{-\infty}^\infty \exp(-x^2) \mathrm{d}x = \frac{\sqrt{\pi}}{2} $.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\expo{-x^{2}}\cos\pars{ax}\,\dd x:\ {\large ?}}$

\begin{align}
&\color{#66f}{\large\int_{0}^{\infty}\expo{-x^{2}}\cos\pars{ax}\,\dd x}
=\half\Re\int_{-\infty}^{\infty}\expo{-x^{2}}\expo{\ic\verts{a}x}\,\dd x
\\[5mm]&=\half\Re\int_{-\infty}^{\infty}
\expo{-\pars{x -\ic\verts{a}/2}^{2} - a^{2}/4}\,\dd x
=\half\,\expo{-a^{2}/4}
\Re\int_{-\infty - \verts{a}\,\ic/2}^{\infty - \verts{a}\,\ic/2}
\expo{-x^{2}}\,\dd x
\\[5mm]&=\half\,\expo{-a^{2}/4}\
\underbrace{\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x}_{\ds{=\ \color{#c00000}{\root{\pi}}}}\ =\
\color{#66f}{\large{\root{\pi} \over 2}\,\expo{-a^{2}/4}}
\end{align}

