Calculate the expected value of $Y=e^X$ where $X \sim N(\mu, \sigma^2)$ I got a problem of calculating $E[e^X]$, where X follows a normal distribution $N(\mu, \sigma^2)$ of mean $\mu$ and standard deviation $\sigma$.
I still got no clue how to solve it. Assume $Y=e^X$. Trying to calculate this value directly by substitution $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\, e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ then find $g(y)$ of $Y$ is a nightmare (and I don't know how to calculate this integral to be honest).
Another way is to find the inverse function. Assume $Y=\phi(X)$, if $\phi$ is differentiable, monotonic, and have inverse function $X=\psi(Y)$ then $g(y)$ (PDF of random variable $Y$) is as follows: $g(y)=f[\psi(y)]|\psi'(y)|$.
I think we don't need to find PDF of $Y$ explicitly to find $E[Y]$. This seems to be a classic problem. Anyone can help?

Let $X$ be an $\mathbb{R}$-valued random variable with the probability density function $p(x)$, and $f(x)$ be a nice function. Then
$$\mathbb{E}f(X) = \int_{-\infty}^{\infty} f(x) p(x) \; dx.$$
In this case, we have
$$ p(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}},$$
hence
$$ \mathbb{E}e^X = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{\infty} e^x e^{-\frac{(x-\mu)^2}{2\sigma^2}} \; dx. $$
Now the rest is clear.
