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

$$\frac{\mathrm{d^2}}{\mathrm{dp^2}} \Gamma(p) = \int_0^{\infty} e^{-x}(\log x)^2x^{p-1}dx$$

  • 1
    $\begingroup$ Dominated convergence theorem. $\Gamma$ is even meromorphic. $\endgroup$ – Yai0Phah Dec 27 '14 at 16:01
  • $\begingroup$ @FrankScience How could I show it is twice continuously differentiable without using that result? $\endgroup$ – user112495 Dec 27 '14 at 16:05

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.


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