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.

The integral relates to the expectation value of the normal hazard function
$$E_X[h(x)] = E_X\left[\dfrac{\phi(x)}{1-\Phi(x)}\right] = \int_{-\infty}^{\infty} \left(\dfrac{\phi(x)}{1-\Phi(x)}\right) \phi(x) dx  = \int_{-\infty}^{\infty} \left(\dfrac{\phi(x)}{\Phi(x)}\right) \phi(x) dx$$
The last step is due to the symmetry.
This hazard function will be approximately asymptotic to $x$ at positive infinity and will be smaller than $x^2$ (except in some region where it is finite).
Since $E_X[x^2]$ is finite so must be $E_X[h(x)]$

The big difference between $x^2$, which is already resulting in a finite integral, and the hazard function $h(x)$ indicates that this is not a hard problem and there are probably many approaches to show that the integral does not diverge (as seen in many answers to the question).
For instance, we can more precisely see this asymptotic behaviour of the hazard function in the expression
$$log(1-F(t)) = log(S(t)) = -\int_{-\infty}^x h(t) dt$$
This can be related to the asymptotic behavior of the error function (as $\text{erfc}(x/\sqrt{2})/2 = 1-F(x)$), which has exponential bounds with factor $e^{-x^2}$ for $x>0$.
