# Intégral of cumulative distribution function and the density function

Let $$\Phi: \mathbb{R} \to \mathbb{R}$$ be the cumulative distribution function of a standard normal variable $$N(0,1)$$ and $$\phi: \mathbb{R} \to \mathbb{R}$$ be its density function. I am interested in evaluating the following integral

$$\int_{0}^{\infty} \Phi(x) \phi(x) \, dx.$$

Note that if the integral is changed to $$\int_{\mathbb{R}} \Phi(x) \phi(x) \,dx$$, then by using the facts that $$\Phi- \frac{1}{2}$$ is an odd function and that $$\phi$$ is an even function,

$$\begin{eqnarray} \int_{\mathbb{R}} \Phi(x) \phi(x) \,dx & = & \int_{\mathbb{R}} \Big( \Phi(x)- \frac{1}{2} \Big) \phi(x) \, dx + \frac{1}{2} \\ & = & \frac{1}{2}. \end{eqnarray}$$

However, this trick does not work in this case when the limits of integration are $$0$$ to $$\infty$$. Integration by substitution (using $$u=-x$$) does not work either. Any ideas?

Intuitively, the answer should be $$\frac{1}{4}$$, by some kind of symmetry. But I can’t show it.

Notice that $$\phi(x) = \Phi^{\prime}(x)$$. Thus if we make the substitution $$u = \Phi(x)$$, we get that $$\mathrm{d}u = \phi(x)\mathrm{d}x$$. Since $$\Phi(0) = 0.5$$ and $$\Phi(\infty) = 1$$, we get \begin{align*} \int_{0}^{\infty}\Phi(x)\phi(x)\mathrm{d}x = \int_{1/2}^{1}u\mathrm{d}u = \frac{1}{2} - \frac{1}{8} = \frac{3}{8} \end{align*}
At your case, we have $$\Phi(-\infty) = 0$$ and $$\Phi(\infty) = 1$$. Consequently, it results that \begin{align*} \int_{\textbf{R}}\Phi(x)\phi(x)\mathrm{d}x = \int_{0}^{1}u\mathrm{d}u = \frac{1}{2} \end{align*}