I'm trying to show that the gamma function is twice continuously differentiable for $p>0$. I was wondering whether the following is actually valid? Or is the simply the way of computing the derivative once you've shown it is twice differentiable. Additionally, how do I then go about showing the second derivative is continuous? Thanks.
$$\Gamma(p) = \int_0^{\infty} x^{p-1}e^{-x} dx$$
By utilising differentiation under the integral sign, we get that
$$\frac{\mathrm{d}}{\mathrm{dp}} \Gamma(p) = \int_0^{\infty} \frac{\partial}{\partial p}(x^{p-1}e^{-x})dx = \int_0^{\infty} e^{-x}(\log x)x^{p-1}dx$$
Differentiating again, we get
$$\frac{\mathrm{d^2}}{\mathrm{dp^2}} \Gamma(p) = \int_0^{\infty} e^{-x}(\log x)^2x^{p-1}dx$$
By the Bohr-Mollerup theorem we know that $\Gamma$ is a log-convex function over $\mathbb{R}^+$.
It is worth to prove that $\log\Gamma\in C^2(\mathbb{R}^+)$, that is equivalent to the continuity of: $$\psi'(x) = \sum_{n=0}^{+\infty}\frac{1}{(x+n)^2}.\tag{1}$$ Over any compact subset of $\mathbb{R}^+$, the RHS of $(1)$ is a totally convergent series of continuous functions, hence a continuous function.