Is upper incomplete gamma function convex?

Question

Considering the definition of upper incomplete gamma function: $\Gamma(a, x) =$ $\int_{x}^{\infty}t^{a-1}e^{-t} dt$

Given that $a$ is fixed and $0 < x < a$, can we prove the function is convex over $x$?

Answer 1

The incomplete gamma function is not convex with respect to x for fixed a for 0 < x < a, as evidenced by the following counterexample.

Counterexample for a = 10, and x = 1, 5, 9:

gammainc(5,10,'upper')
    0.9682
0.5*(gammainc(1,10,'upper')+gammainc(9,10,'upper'))
    0.7937

Answer 2

No, you cannot prove it is convex in $x$ because it is not, as Mark's counterexample indicates. For fixed $a$, $\Gamma$ is concave for $x\in (0, a-1)$ and convex for $x\in (a-1, a).$ Differentiating twice tells the story: $$\frac{\partial \Gamma}{\partial x} = -x^{a-1} e^{-x}$$and $$\frac{\partial^2 \Gamma}{\partial x^2} = e^{-x} x^{a-2} (x - a + 1).$$ The sign of $x-a+1$ determines the sign of the second partial derivative.