Hello,


Here is an interesting problem. It looks elementary, but it has taken me some efforts without solving it. Let


$$
h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int\_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy.
$$


The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function? 


It seems a quite easy problem. By taking the first derivative, we need to prove that


$$
h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}} \ge 0.
$$
which again, not obvious (for $x<0$). Some facts, that might be useful, are:


$$
\lim\_{x\rightarrow -\infty} h(x) =0, \quad \lim\_{x\rightarrow -\infty} h(x)' =0.
$$


Thank you very much for any hints!


Anand



Is the function $e^{x^2/2} \Phi(x)$ monotone increasing?
We can write $h(x)=\frac 1{\sqrt{2\pi}}\int\_{-\infty}^x \exp\left(\frac{x^2-y^2}2\right)dy$. Now put $t=x-y$. We get 
\begin{align}
h(x)&=\frac 1{\sqrt{2\pi}}\int\_0^{+\infty}\exp\left(\frac{x^2-(x-t)^2}2\right)dt\\\
&=\frac 1{\sqrt{2\pi}}\int\_0^{+\infty}\exp\left(xt-\frac{t^2}2\right)dt.
\end{align}
We can differentiate under the integral thanks to the dominated convergence theorem. We get 
$$h'(x)=\frac 1{\sqrt{2\pi}}\int\_0^{+\infty}t\exp\left(xt-\frac{t^2}2\right)dt\geq 0.$$
Added later: we don't need to diffentiate. If $x\_1\leq x\_2$ then for $t\geq 0$ we have $e^{tx\_1}\leq e^{tx\_2}$ therefore $ h(x\_1)\leq h(x\_2) $.


