Say $Y=Log_2[1+x]=g(X)$ and $f_X = \frac{e^{-\frac{(\mu -\log (x))^2}{2 \sigma ^2}}}{\sqrt{2 \pi } x \sigma }$ is Log-normal density function: [Wiki]
Find E[Y]?
Since $E[Y] = \int_0^\infty y f_Y \ dy = \int_0^\infty g(x)f_X(x)dx$. So, I do not need to find $f_Y(y)$.
$$\int_0^{\infty } \frac{\log_2 (1+x) e^{-\frac{(\mu -\log (x))^2}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma x} \, dx$$
Unfortunately, I'm having difficulty proceeding from here.
I tried solving this with Mathematica functions but it also has no idea how to solve it. Will appreaciate any help on the evaluation steps.
Replace x with $x+1$ in the expression of $f_{_X}$, and factor all constants outside the integral. Then use
$$\int_1^\infty\frac{\ln x}x\cdot\exp\bigg[\bigg(\frac{\mu-\ln x}a\bigg)^2~\bigg]dx~=~\frac{a^2}2\cdot\exp\bigg[-\bigg(\frac\mu a\bigg)^2~\bigg]+\dfrac{\sqrt\pi}2a\mu\cdot\bigg[1+\text{erf}\bigg(\dfrac\mu a\bigg)\bigg].$$
More to the point, $\displaystyle\int_0^\infty\frac{\ln x}x\cdot\exp\bigg[\bigg(\frac{\mu-\ln x}a\bigg)^2~\bigg]dx~=~a\mu~\sqrt\pi$ , where the quantity on $(0,1)$
is negligible. Obviously, $a=\sigma\sqrt2,~$ and $~\log_ut=\dfrac{\ln t}{\ln u}~=>~\log_2(x+1)=\dfrac{\ln(x+1)}{\ln2}$.