How to compute this integral involving a cdf?

Consider $I(a) = \int_0^\infty \Phi(a x) \mathrm{d} \Phi(x)$. Differentiate with respect to $a$, and denote $\phi(x) = \Phi^\prime(x)$: $$ I^\prime(a) = \int_0^\infty x \phi(a x) \phi(x) \mathrm{d} x = \frac{1}{2 \pi} \int_0^\infty x \mathrm{e}^{-\frac{(1+a^2) x^2}{2}} \mathrm{d} x = \frac{1}{2 \pi} \frac{1}{1+ a^2} $$ Now, noting that $I(0) = \int_0^\infty \frac{1}{2} \mathrm{d} \Phi(x) = \frac{1}{4}$: $$ I\left(a\right) = \frac{1}{4} + \frac{1}{2 \pi} \int_{0}^{a} \frac{\mathrm{d} a}{1+a^2} = \frac{1}{4} + \frac{1}{2 \pi} \arctan(a) $$ Now $I\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{4} - \frac{1}{2 \pi} \arctan\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2 \pi} \arctan\left(\sqrt{2}\right) \approx 0.152043 $

Sasha's comment following his answer suggests a different calculation that does not require knowledge of the antiderivative of $(1+a^2)^{-1}$, only pie-cutting or using the circular symmetry of the joint density of two independent standard normal random variables . $$\begin{align*} \int_0^{\infty}\Phi(ax)\;\mathrm d\Phi(x) &= \int_0^{\infty}\Phi(ax)\phi(x)\mathrm\; dx\\ &= \int_0^{\infty}\left[ \int_{-\infty}^{ax}\phi(y)\;\mathrm dy\right] \phi(x)\mathrm\; dx\\ &= \int_0^{\infty} \int_{-\infty}^{ax}\phi(y) \phi(x)\;\mathrm dy\;\mathrm dx\\ &= \frac{1}{2\pi}\int_{r=0}^{\infty}\int_{\theta=-\pi/2}^{\arctan(a)} \exp(-r^2/2) \cdot r\;\mathrm d\theta\;\mathrm dr\\ &= \frac{\arctan(a)+\pi/2}{2\pi}\\ &= \frac{1}{4} + \frac{1}{2\pi}\arctan(a). \end{align*}$$