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

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 convex 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