How to show that this integral of the normal distribution is finite? Numerically, I have noticed that
$$\int_{-\infty}^{\infty} \dfrac{\phi(x)^2}{\Phi(x)}dx < \infty$$
where $\phi$ and $\Phi$ are the standard normal pdf and cdf. However, I do not see how to prove it. I would appreciate any hints.

Intuitive approach
$\Phi(x)$ is changing in the interval $[0,1]$, whereas $\phi(x)^2$ is itself a normal distribution which is integrable. Thus, the only reason why the integral could not be finite, is because it diverges in the limit $x\rightarrow -\infty$.
By L'Hôpital's rule:
$$
\frac{\phi^2(x)}{\Phi(x)}\sim
\frac{2\phi(x)\phi'(x)}{\phi(x)}\sim
\phi'(x)= -\frac{2x}{\sigma^2}\phi(x)\\(\text{as }x\rightarrow -\infty)
$$
In other words, the asymptotic form of the integrand for $x\rightarrow-\infty$ is
$$
\frac{\phi^2(x)}{\Phi(x)}\sim -\frac{2x}{\sigma^2}\phi(x),
$$
the integral of which converges in the desired limit.
Rigorous approach
Perhaps, a more rigorous way could be to show that in the limit $x\rightarrow -\infty$ the integrand decays faster than $1/x$,a nd hence convergent.
We this apply the L'Hôpital's rule to the following:
By L'Hôpital's rule:
$$
\lim_{x\rightarrow -\infty}\frac{x\phi^2(x)}{\Phi(x)}=
\lim_{x\rightarrow -\infty}\frac{\phi^2(x)+x\phi(x)\phi'(x)}{\phi(x)}=
\lim_{x\rightarrow -\infty}\left[\phi(x)+x\phi'(x)\right]=
\lim_{x\rightarrow -\infty}\left[1-\frac{x}{\sigma^2}\right]\phi(x)=0.
$$Thus, the integral converges in the limit in question.
Remarks:

*

*I appreciate @SextusEmpiricus remarks, which led to improvement of this answer.

*$$\phi(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$$
