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Say $Y=g(X)$ and $p_X = \frac{e^{-\frac{(\mu -\log (x))^2}{2 \sigma ^2}}}{\sqrt{2 \pi } x \sigma }$ 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 $f_Y(y)$.

.    
px = Simplify[PDF[LogNormalDistribution[Mu, Sigma], x], x > 0]
Integrate[Log2[1 + x] px, {x, 0, \Infinity}, Assumptions -> {Sigma > 0}]
.

Unfortunately, Mathematica cannot evaluate this integral for a closed form expression and I'm not quite sure how else to solve it. Will appreaciate any help I can get on the evaluation steps.

Edit:

I've also tried the code below with no results:

 Expectation[Log2[1 + x], x \[Distributed] LogNormalDistribution[Mu, Sigma]]
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2 Answers 2

I suspect there may not be a closed form expression (I have not looked at it hard enough). If the aim is to not analytical but numerical, I post the following for illustration (apologies if not the intent of question):

f[x_, y_] := 
 NIntegrate[
  Log2[t + 1] Exp[-(x - Log[t])^2/(2 y^2)]/(Sqrt[2 Pi] t y), {t, 0, 
   Infinity}]
rv[a_, b_, n_] := 
 Mean[Log2[1 + RandomVariate[LogNormalDistribution[a, b], n]]]
Manipulate[
 Show[Plot[f[m, s], {m, 1, 5}, PlotRange -> {0, 7}], 
  ListPlot[Table[{j, rv[j, s, n]}, {j, 1, 5, 0.1}], 
   PlotMarkers -> {"\[FilledDiamond]", 10}, PlotStyle -> Red]], {s, 1,
   5}, {n, {10, 100, 1000, 10000}}]

enter image description here

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You can also depict the expected value nicely as a function of Mu and Sigma using Plot3D.

px = Simplify[PDF[LogNormalDistribution[Mu, Sigma], x], x > 0]

E^(-((Mu - Log[x])^2/(2 Sigma^2)))/(Sqrt[2 \[Pi]] Sigma x)

Plot3D[NIntegrate[Log2[1 + x] px, {x, 0, \[Infinity]}], {Mu, -5, 
  10}, {Sigma, .2, 2}, AxesLabel -> {Mu, Sigma, EY}]

Expected value of a log-normal Distribution as a function of Mu and Sigma

PS: Reminds me of Black-Scholes formula for the evaluation of an Option.

Extending the range of Sigma Shows the nontrivial dependence more clearly:

Expected value of a log-normal Distribution as a function of Mu and Sigma

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