How to prove $\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}$ I've been grappling for a few hours with this problem ($b$ is a parameter)
$\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}$
Tried direct integration by parts, differentiation w.r.t. $b$, then integrating... all to no avail.
I think the second approach is the way... Pls help
 A: Contour Integral Method
$$
\begin{align}
\int_0^\infty e^{-x^2}\cos(2bx)\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty e^{-x^2}\cos(2bx)\,\mathrm{d}x\\
&=\frac12\int_{-\infty}^\infty e^{-x^2}e^{i2bx}\,\mathrm{d}x\\
&=\frac12\int_{-\infty}^\infty e^{-(x-ib)^2}e^{-b^2}\,\mathrm{d}x\\
&=\frac12e^{-b^2}\int_{-\infty-ib}^{\infty-ib}e^{-x^2}\,\mathrm{d}x\\
&=\frac12e^{-b^2}\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\tag{$\ast$}\\
&=\frac{\sqrt\pi}{2}e^{-b^2}
\end{align}
$$
$(\ast)$ is valid since $e^{-z^2}$ has no poles and the integral along the two ends of the infinitely long rectangle between the paths vanishes at $\infty$.

Differential Equation Method
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}b}\int_0^\infty e^{-x^2}\cos(2bx)\,\mathrm{d}x
&=-\int_0^\infty e^{-x^2}2x\sin(2bx)\,\mathrm{d}x\\
&=\int_0^\infty\sin(2bx)\,\mathrm{d}e^{-x^2}\\
&=-2b\int_0^\infty e^{-x^2}\cos(2bx)\,\mathrm{d}x
\end{align}
$$
The solution to $\frac{\mathrm{d}}{\mathrm{d}b}f(b)=-2bf(b)$ is $f(b)=Ce^{-b^2}$. Evaluating at $b=0$ yields $C=\frac{\sqrt\pi}{2}$. Therefore,
$$
\int_0^\infty e^{-x^2}\cos(2bx)\,\mathrm{d}x=\frac{\sqrt\pi}{2}e^{-b^2}
$$
