Show that the second derivative $\Gamma''(x)$ is positive when $x>0$

Question

Let $\Gamma(x)=\int_0^{\infty}t^{z-1}e^{-t}dt$.

I know that the first derivative is positive, since $\Gamma(x)$ is increasing when $x>0$, but I don't know how to show that the second derivative is positive without calculating it, something which we have not yet learned to do. Plotting out $x!$ shows that $\Gamma$ is concave up, and has a positive second derivative, but I don't know how to formulate a formal proof.

Would it suffice to say $\Gamma(x+2)-\Gamma(x+1)>\Gamma(x+1)-\Gamma(x)$, $\forall x>0$, and $\Gamma$ is increasing at an increasing rate?

Answer 1

If you define $$\Gamma(x) = \int_0^\infty t^{x-1} e^{-t}\; dt$$ then $$\Gamma''(x) = \int_0^\infty t^{x-1} \ln(t)^2 e^{-t}\; dt > 0$$

BTW, it's not true that $\Gamma'(x) > 0$ for $x > 0$. $\Gamma$ has a minimum at approximately $1.4616321449683623413$.

Answer 2

We know that $\log\Gamma(x)$ is convex on $\mathbb{R}^+$ due to the Bohr-Mollerup theorem.

So $\Gamma(x)$ is a convex function on $\mathbb{R}^+$ as the exponential of a convex function, and $\Gamma''(x)\geq 0$.