# How demonstrate the Craig representation for the Gaussian probability function?

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.

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Thanks @DavideGiraudo for fixing it. – jack May 27 '12 at 20:50
here a way to solve this kaust.academia.edu/AhmadMawla/Papers/513176/Q-Function – user33125 Jun 7 '12 at 0:46

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.