Let $$\mu_x=\frac{1}{\Gamma(x+1)}\int_0^{\infty}u^x f(u) du \tag{1}$$

Suppose that $f(u)>0$ when $u>0$ and $f(u)\to 0$ fast enough when $u\to\infty$ so that $\mu_x,-1<x<\infty$ converges.

It is known (Ref.1) that if $\log f(u)$ is concave on $(0,\infty)$, then $\mu_x$ is concave on $(-1,\infty)$

Question: Let $$\nu_x=\frac{1}{G(x+1)}\int_0^{\infty}u^x f(u) du \tag{2}$$

Dose there exist a function $G(x)>0$ such that if $\log f(u)$ is concex on $(0,\infty)$, then $\nu_x$ is convex on $(-1,\infty)$?

The function $G(x)$ I am searching for might look like $\Gamma(ax+b)$ where $a,b$ are two parameters indepedent of $x$.

Ref. 1 TURAN INEQUALITIES AND ZEROS OF DIRICHLET SERIES ASSOCIATED WITH CERTAIN CUSP FORMS J. B. CONREY AND A. GHOSH transactions of the american mathematical society Volume 342, Number 1, March 1994