Prove that an elementary function is non-decreasing For $p > 0$, define : 
$f(p) := \int_0^1{p (1-x)^{p^2 + 2p}e^{\frac{(1+p)^2}{2} x(x+2)}}dx$
The question is : how to prove that $f$ is a non-decreasing function of p?
It seems to be the case numerically.
Note that one can show that $p \mapsto f(p)/p$ is a non-increasing function because $f(p)/p = \int_0^1{\frac{1}{1-x}e^{(1+p)^2[\ln{(1-x)} + \frac{x(x+2)}{2}]}dx}$ and $\ln{(1-x)} + \frac{x(x+2)}{2}$ is always negative. So the difficulty here is to understand why does the factor $p$ help!
Thanks a lot for any clue!
 A: It isn't difficult, just somewhat tricky.
Let $w(x)=\log(1-x)+x+\frac{x^2}2=-\frac{x^3}3-\frac{x^4}4-\dots$. We need to show that the product of $f(p)=\int_0^1\frac 1{1-x}e^{(1+p)^2w(x)}\,dx$ and $p$ is increasing. Note that $-w'(x)=\frac 1{1-x}-1-x$, so $f(p)=\frac 1{(1+p)^2}+\int_0^1(1+x)e^{(1+p)^2w(x)}\,dx=\frac{1}{(1+p)^2}+g(p)$. 
Claim 1: The function $u(p)=(p+1)g(p)$ is increasing.
Indeed, take any $q>p$ and define $a=\frac{1+p}{1+q}$. Notice now that $w$ is decreasing and $(1+q)^2 w(a^{2/3}x)\ge (1+p)^2 w(x)$, so it suffices to show that for every $y\in(0,1)$, we have
$$
\int_0^{a^{2/3}y}(1+x)\,dx=a^{2/3}y+a^{4/3}\frac{y^2}2\ge a\int_0^y(1+x)\,dx=ay+a\frac{y^2}2
$$ 
But since $\frac{y^2}2<y$, the two terms on the left have the same product as the two terms on the right, but are farther apart, so this is obvious.
Claim 2: $u(p)\ge 1$. 
Indeed, we have
$$
u(p)\ge u(0)=g(0)=\int_0^1(1-x^2)e^{x+\frac{x^2}2}\,dx\ge\int_0^1(1-x^2)(1+x+x^2)\,dx
\\
=1-\frac 13+\frac 12-\frac 14+\frac 13-\frac 15=1+\frac 14-\frac 15>1\,.
$$
Now it is time to differentiate $pf(p)=\frac{p}{(1+p)^2}+u(p)\frac{p}{1+p}$. We get 
$$
\frac 1{(1+p)^2}-\frac{2p}{(1+p)^3}+\frac{u(p)}{(1+p)^2}+u'(p)\frac{p}{1+p}> \frac{1+u(p)-2}{(1+p)^2}+u'(p)\frac{p}{1+p}\ge 0\,.
$$
The end.
