Here's what I did:
$$\Gamma'(z)=\int_0^\infty \log(t)e^{-t}t^{z-1}dt$$
$$\Gamma''(z)=\int_0^\infty \log^2(t)e^{-t}t^{z-1}dt$$$$\frac{d^2}{dz^2}\log\Gamma(z)=\frac{\Gamma''(z)\Gamma(z)-(\Gamma'(z))^2}{\Gamma^2(z)}$$
If denote $(e^{-t}t^{z-1})^{1/2}$ as $a$, $(\log^2(t)e^{-t}t^{z-1})^{1/2}$ as $b$, then by cauchy-schwarz inequality, $(\int ab)^2\le(\int a^2)(\int b^2) $, the second derivative of $\log\Gamma$ is greater than zero. Am I right? I checked digamma and polygamma function, but their expressions look weird.