TITLE: How demonstrate the Craig representation for the Gaussian probability function?
QUESTION [7 upvotes]: The Q-function is defined by : 
$$Q(x) =\frac{1}{\sqrt{2\pi}} \int_{x}^{\infty}\exp(-\frac{u^2}{2}) \ \mathrm{d}u \ \ (1).$$
According to the wiki page there is an alternative form of the Q-function based on John W. Craig's work  that is more useful is expressed as:
$$Q(x) =\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\exp\left(-\frac{x^2}{2\sin^2(\theta)}\right) \ \mathrm{d}\theta  \ \ (2).$$ 
Craig's proove is based on probabilistic  approach, there for I look for an analytic one.
any help will be appreciated.
Thanks.

REPLY [4 votes]: Since both expressions coincide at $x=0$, it suffices to show that their derivatives coincide on $x\geqslant0$. Since the derivative of the first expression of $Q(x)$ is proportional to $\mathrm e^{-x^2/2}$, this happens if $R(x)$ is constant on $x\geqslant0$, where
$$
R(x)=\mathrm e^{x^2/2}\int_0^{\pi/2}\frac{x}{\sin^2\theta}\mathrm e^{-x^2/(2\sin^2\theta)}\mathrm d\theta=\int_0^{\pi/2}\frac{x}{\sin^2\theta}\mathrm e^{-x^2\cot^2\theta/2}\mathrm d\theta.
$$
The change of variables $v=x\cot\theta$ yields the range $v\gt0$ and the Jacobian $\mathrm dv=x\mathrm d\theta/\sin^2\theta$, hence
$$
R(x)=\int_0^{+\infty}\mathrm e^{-v^2/2}\mathrm dv,
$$
which does not depend on $x$. QED.