Given $X \sim \operatorname{Gamma}(\kappa, \theta)$ with CDF $F_X(\kappa, \theta)$, where $\kappa \geq 1$ and $\theta > 0$, the expected value of $\mathbb{E} \left\{ \ln(1+X) \right\}$ is calculated by $\int_{x} \ln(1+x) \, dF_X(\kappa, \theta)$.

For calculation simplification, we want to get its approximated expression $f(\kappa, \theta)$ with $\mathbb{E} \left\{ \ln(1+X) \right\} \geq f(\kappa, \theta)$.

One example is to set $f(\kappa, \theta) = \ln \left(1 + \alpha \mathbb{E} \left\{ X \right\} \right)$, and an approximated solution is to set $\alpha = \left( e^{\psi(\kappa) + \ln(\theta)} - 1 \right) / \kappa \theta$, such that

$$ f\left(\kappa,\theta\right)=\ln\left(1+\alpha\mathbb{E}\left\{ X\right\} \right)=\psi\left(\kappa\right)+\ln\left(\theta\right)=\mathbb{E}\left\{ \ln\left(X\right)\right\} \le\mathbb{E}\left\{ \ln\left(1+X\right)\right\}. $$

However, this approximation is too loose when $\theta$ is small.

In summary, we want to get a tighter approximation, for example, finding the function of $\alpha$ over $\theta$ and $\kappa$, as shown in the figure.