Assuming I have a Gamma distributed random variable $X\sim\Gamma(\alpha,1)$, $\alpha>1$. $E(X)=\alpha$.
$Y=\ln(1+bX)$, $b$ is a constant.
I like to have a closed form expression of the following expectation value (integrals): \begin{eqnarray*} E(Y)=E(\ln(1+bX)) &=& \int_0 ^\infty \ln(1+bx)f(x)\,\mathrm{d}x \\&=& \int_0 ^\infty \ln(1+bx）\frac{x^{\alpha-1}e^{-x}}{\Gamma(\alpha)}\,\mathrm{d}x \end{eqnarray*} I don't know how to get the closed and simplest form of this value $E(Y)$.
Could any expert help me? Thank you very much.