x is a normal distributed variable. then what is the expectation of ln(1+e^x).


i simulated this distribution and find that when x is N(0, 100), the mean of this function is around 4.1, and when x is N(-10, 100), the mean is around 0.9. 


can anyone tell me the exact expectation of this function, when x is normally distributed?



expectation of ln(1+e^x)
It is easy to obtain lower bounds on the expectation by noting that 
$$
 \ln(1+e^x) \geq \max\{0,x\} 
$$
which simplifies the expectation extremely and leads to the result for $X\sim \mathcal{N}(\mu,\sigma)$
$$
 E(\ln(1+\exp(X))) \geq E(\max\{0,X\}) = \frac{\sqrt{\sigma}}{\sqrt{2\pi}} e^{-\frac{\mu^2}{2\sigma}} + \mu \Phi\_{\mu,\sigma}(0) ,
$$
where $\Phi\_{\mu,\sigma}$ is the distribution function of $X$. 


The bounds obtained by this formula are $\mathcal{N}(0,100) : 3.99$ and $\mathcal{N}(-10,100): 0.833$.


