integration - How to show that $intlimits_{-infty}^{+infty}(n-1)Phi(x)^{n-2}phi(x)^2dx$? decreases in $n$?

I was working on a research project that involves taking the integral of

$$(n-1)\int\limits_{-\infty}^{+\infty} \Phi\left(x\right)^{n-2}\phi\left(x\right)^2dx,$$ where $\Phi(.)$ is the CDF for standard normal, $\phi$ the PDF, $\alpha>0$ and $n\in\mathbb{Z}$ and $n>3$.

Eventually, I wish to show that the whole expression decreases monotonically as $n$ increases.

Any help on this will be greatly appreciated. Thanks!

Answer

An integration by parts using $u=\varphi$ and $v'=(n-1)\varphi\Phi^{n-2}$, hence $u'(x)=-x\varphi(x)$ and $v=\Phi^{n-1}$, shows that the $n$th term is

$$A_n=\int_\mathbb Rx\varphi(x)\Phi(x)^{n-1}\mathrm dx=\int_0^\infty x\varphi(x)B_n(\Phi(x))\mathrm dx,$$ where $$B_n(t)=t^{n-1}-(1-t)^{n-1}.$$
For every $t$ in $(\frac12,1)$ and every $n\geqslant3$, $$B_n(t)-B_{n-1}(t)=t(1-t)((1-t)^{n-3}-t^{n-3})\leqslant0,$$ hence $A_n\leqslant A_{n-1}$, that is, the sequence $(A_n)_{n\geqslant2}$ is non increasing.