Is $\Gamma(\alpha, 0, x)$ log-concave as a function of $x$ (for fixed $\alpha$)? The incomplete Gamma function is defined as:
$$\Gamma(\alpha, 0, x) = \int_0^x t^{\alpha - 1} e^{-t} dt$$
Let $\alpha \ge 1$ be a fixed number. Is $\Gamma(\alpha, 0, x)$ log-concave, as a function of $x$?
Assume that $x\ge 0$.
P.S. A function is log-concave if its logarithm is concave. 
 A: According to your definition,
$$ \frac{d}{dx}\Gamma(\alpha,0,x) = x^{\alpha -1}e^{-x},\qquad \frac{d^2}{dx^2}\Gamma(\alpha,0,x) = x^{\alpha-2}e^{-x}(\alpha-1-x).$$
Let we just set $f(x)=\Gamma(\alpha,0,x)$ in order to have a more efficient notation. We may show that $f$ is a log-concave function by showing that $\frac{d^2}{dx^2}\log f(x)<0$, i.e.
$$ f''(x)\,f(x) \leq f'(x)^2 $$
that is equivalent to:
$$ x^{\alpha-2}e^{-x}(\alpha-1-x)\int_{0}^{x}t^{\alpha-1}e^{-t}\,dt \leq x^{2\alpha-2}e^{-2x} $$
or to:
$$ (\alpha-1-x)\int_{0}^{x}t^{\alpha-1}e^{-t}\,dt \leq x^{\alpha}e^{-x} $$
or to:
$$ (\alpha-1-x)\int_{0}^{1}(1-u)^{\alpha-1}e^{xu}\,dt \leq 1\tag{1}$$
that is trivial if $x\geq (\alpha-1)$, and is a simple consequence of $(1-u)\leq e^{-u}$ otherwise.
Hence yes, $f$ is a log-concave function.

Interestingly, $g(\alpha)=\Gamma(\alpha,0,x)$ is a log-convex function with respect to $\alpha$ due to the Cauchy-Schwarz inequality, since $\frac{d^k}{d\alpha^k}\Gamma(\alpha,0,x)=\int_{0}^{x}\left(\log t\right)^k t^{\alpha-1}e^{-t}\,dt$.
