$\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$ is the pdf of a standard normal distribution.


$\Phi(x) = \int\_{- \infty}^x \phi(t) dt$ is the cdf of a standard normal distribution.


How does one calculate the following: 
$$
T = \int\limits\_0^{\infty} \Phi^2(bx) \phi(x) dx 
$$


In fact, I can find the result in wiki,
<https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions#CITEREFOwen1980>


But I would like to know how to do it.



Integral of product of Gaussian pdf and cdf
Maybe this proof is too large, but you can deduce the result you want, sorry for that.


First of all:


* $\phi(x)= \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$
* $\Phi(x)=\int\_{-\infty}^{x}\phi(t)dt=\frac{1}{2}(1+erf(\frac{x}{\sqrt{2}}))$


To prepare the proof, start with this facts:


1. $\int\_{0}^{\infty}xe^{-x^2a}dx=\frac{1}{2a}$ (Proof by direct integral)
2. $erf'(x)=\frac{2}{\sqrt{\pi}}e^{-x^2}$ (Proof by definition of $erf$ function)
3. $$\int\_0^{\infty}xe^{-\alpha^2x^2}erf(\beta x)dx=\frac{\beta}{2\alpha^2 \sqrt{\alpha^2+\beta^2}} $$
**Hint for 3.**
Integrating by parts, with 
\begin{equation}u=erf(\beta x) \rightarrow du=\frac{2\beta}{\sqrt{\pi}}e^{-(x\beta)^2}dx\end{equation}
$$dv=xe^{-\alpha^2x^2} \rightarrow v=-\frac{e^{-x^2\alpha^2}}{2\alpha^2}$$
the result follows really easy.


Now, you have all the ingredients to calculate this monster:
$$T(b)=\int\_{0}^{\infty}\Phi^2(bx)\phi(x)dx$$
Derivating $T(b)$ respect $b$, changing $\Phi(bx)$ by the definition in terms of $erf$ function, and using the fact 2, will lead you to:


$$T'(b)=...=\frac{1}{2\pi}\Big(\int\_0^{\infty}{xe^{-x^2(b^2+1)/2}}dx + \int\_0^{\infty}{x \cdot erf\big(\frac{bx}{\sqrt{2}}\big)e^{-x^2(b^2+1)/2}}dx \Big) =\frac{1}{2\pi}(A+B)$$


Compute A using the fact 1 to get:
$$A=\frac{1}{b^2+1}$$
and compute B using the fact 3 with $\alpha=\sqrt{\frac{b^2+1}{2}}$ and $\beta= \frac{b}{\sqrt{2}}$ to get:
$$B=\frac{b}{(b^2+1)\sqrt{2b^2+1}}.$$


It remains to integrate $T'(b)=\frac{A+B}{2\pi}$ respect $b$:
\begin{align}T(b)=&\frac{1}{2\pi}\Big(\int\_{0}^{b}\frac{1}{t^2+1}dt+\int\_{0}^{b}\frac{t}{(t^2+1)\sqrt{2t^2+1}}dt\Big)=\\
=&\frac{1}{2\pi}\Big(arctan(b)+arctan\big(\sqrt{2b^2+1}\big)\Big)+C. \hspace{0.5cm}\square
\end{align}


