analysis - Gaussian-like integral : $int_0^{infty} e^{-x^2} cos( a x) mathrm{d}x$

Prove that :

$$\frac{\sqrt{\pi}}{2} e^{-\frac{a^2}{4} } =\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$$

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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 ?

Answer

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 $f(0)=\int_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}2$,
$$\int_0^\infty e^{-x^2}\cos(ax)\,\mathrm{d}x=\frac{\sqrt\pi}{2}\,e^{-a^2/4}\tag3$$

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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 $\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}t=\sqrt\pi$