Show that function is logarithmically concave Let $p: [0,\infty)\rightarrow [0,\infty)$ be a strictly increasing convex function.
Let $T$ be defined as $$T(x)=\int_x^\infty \frac{1}{e^{p(t)}} dt$$
We are to show that $\ln T$ is concave.
I've managed to show that $$T'(x)=-\frac{1}{e^{p(x)}}$$
that $T$ is convex and I've tried using the fact that $$\int_a^{\frac{a+b}{2}} \frac{1}{e^{p(t)}}dt \geq \int_{\frac{a+b}{2}}^{b} \frac{1}{e^{p(t)}}dt$$
for any $a,b\in[0,\infty)$, but so far has gotten nowhere really. It does seem to be the case judging by simple examples of $p$ like polynomials.
I would appreciate any hints.
 A: Just extending on your line of thought :
$g = ln(T)$
$g' = \frac{1}{T}T'$
$g'' = \frac{1}{T}T'' + T'\frac{-1}{T^{2}}$
$g'' = \frac{TT'' - (T')^{2}}{T^{2}}$
Now, we want to show,
$TT'' - (T')^{2} < 0 \implies e^{-p(x)}p'(x)\int_x^\infty {e^{-p(t)}} dt < e^{-2p(x)}$
We can do relevant division, because $e^{-p(x)} > 0$, to give us
$p'(x)\int_x^\infty {e^{-p(t)}} dt < e^{-p(x)}$
Let us take, $s(x) = e^{-p(x)} - p'(x)\int_x^\infty {e^{-p(t)}} dt$
$s'(x) = -e^{-p(x)}p'(x) - p'(x)(-e^{-p(x)}) - p''(x)\int_x^\infty {e^{-p(t)}} dt$
$s'(x) = -p''(x)\int_x^\infty {e^{-p(t)}} dt$
Since $p$ is convex, $p''(x) > 0$. Since we are integrating a positive value, $\int_x^\infty {e^{-p(t)}} dt > 0$ for all $x$ in domain.
So, $s'(x) < 0$, i.e., $s(x)$ is decreasing.
Now, we find $lim_{x \to \infty} s(x) = lim_{x \to \infty} e^{-p(x)} - p'(x)\int_x^\infty {e^{-p(t)}} dt$.
As you have observed already, $p$ is increasing, and all values of $p$ are positive, so $p \to \infty$ as $x \to \infty$. Hence, $e^{-p(x)} \to 0$ as $x \to \infty$.
Because of the upper limit of the integral being $\infty$, that will go to 0 as well, as $x \to \infty$.
So, $lim_{x \to \infty} s(x) = 0$. This combined with the fact that $s$ is decreasing, we have, $s(x) > 0$, that is, $e^{-p(x)} - p'(x)\int_x^\infty {e^{-p(t)}} dt > 0$.
Hence, $g''(x) < 0$.
