Clever way of finding $\int_0^\infty x\Phi(x)\phi(x)\,\mathrm{d}x$

Question

Suppose that $\Phi$ and $\phi$ are the Standard Normal c.d.f and p.d.f. respectively. Then, evaluate $$\int_0^\infty x\Phi(x)\phi(x)\,\mathrm{d}x$$

There is no use of my trying to show my approach because none of the techniques I used could initiate the solving process. I have, however, obtained the value of $$\int_{-\infty}^\infty x\Phi(x)\phi(x)\,\mathrm{d}x=\dfrac1{\sqrt{2\pi}}$$Any help is appreciated.

The way I proceeded:

1. Orthogonalization of the unit square in which the points $(X,Y)$ are situated. This did not work.

2. Supposing that the centroid of the triangle with vertices at $(0,0),(0,z),(z,0)$ will be the answer due to the symmetry between $X$ and $Y$.

3. Integration by parts repeatedly taking $\Phi$ as the second function, which yielded nothing practically.

4. Using the conditional m.g.f.

The question arose in an attempt to find $\mathbb{E}(X+Y|X>0,Y>0)$ where one term contained this integral. The question came in a semester exam. I will not add "self-study" to this because the question precisely stated that this is an immensely important problem in Monte Carlo Methods, something which I have not read.

Answer 1

Note that $$ -xe^{-x^2/2}\,\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}x}e^{-x^2/2} $$ Since $\Phi(0)=\frac12$, we can integrate by parts to get $$ \begin{align} \int_0^\infty x\Phi(x)\phi(x)\,\mathrm{d}x &=\frac1{\sqrt{2\pi}}\int_0^\infty x\Phi(x)e^{-x^2/2}\,\mathrm{d}x\\ &=-\frac1{\sqrt{2\pi}}\int_0^\infty \Phi(x)\,\mathrm{d}e^{-x^2/2}\\ &=\frac1{2\sqrt{2\pi}}+\frac1{\sqrt{2\pi}}\int_0^\infty e^{-x^2/2}\left(\frac1{\sqrt{2\pi}}e^{-x^2/2}\right)\,\mathrm{d}x\\ &=\frac1{2\sqrt{2\pi}}+\frac1{2\pi}\int_0^\infty e^{-x^2}\,\mathrm{d}x\\ &=\frac1{2\sqrt{2\pi}}+\frac1{2\pi}\frac{\sqrt\pi}2\\ &=\frac{1+\sqrt2}{4\sqrt\pi} \end{align} $$

Answer 2

Have you tried integration by parts together with the knowledge that the antiderivative of $x\phi(x)$ is $-\phi(x)$? The answer is readily computable via this approach.