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.

Here is a self-contained elementary argument, by a comparison with the Laplace distribution. We show
$$\int_{-\infty}^{\infty}\frac{\phi(x)^2}{\Phi(x)}dx<\frac{1}{2\sqrt{\pi}}+\sqrt{\frac\pi2}\simeq 1.535<\infty$$
The positive side of the integral has the bound
$$\int_{0}^{\infty}\frac{\phi(x)^2}{\Phi(x)}dx<
\int_{0}^{\infty}\frac{\phi(x)^2}{1/2}dx=\frac{1}{2\sqrt{\pi}}$$
The negative side of the integral has the bound
$$\int_{-\infty}^{0}\frac{\phi(x)^2}{\Phi(x)}dx<
\int_{-\infty}^{0}\sqrt{2\pi}\frac{f_L(x)^2}{F_L(x)}dx=\sqrt{\frac\pi2}\phantom{1<\infty}$$
Since the Laplace distribution has $f_L(x)=F_L(x)=\frac12 e^x$ on $x<0$, the integral is easy.
To prove that $\phi^2/\Phi<\sqrt{2\pi} f_L^2/F_L$, we start by proving
$$g(u)=\frac{-(2u+1)\exp\left(\frac{-u^2}2-u\right)}{\pi}<1$$

Using the first-derivative test, the maximum occurs at exactly $u_\max=-(3+\sqrt{17})/4$, and by numerical evaluation $g(u_\max)<1$. Thus for $x<0$,
\begin{align}
\phi(x)^2\frac{F_L(x)}{f_L^2(x)}
&=\frac{\exp(-x^2)}{2\pi}\frac{\frac12 \exp(x)}{\frac14 \exp(2x)}\\
&=\frac{\exp(-x^2-x)}{\pi}\\
&=\int_{u=-\infty}^{x}\frac{-(2u+1)\exp(-u^2-u)}{\pi}du\\
&<\int_{u=-\infty}^{x}\exp(-u^2/2)du\\
&=\sqrt{2\pi}\Phi(x)
\end{align}
which is what we need.
