find a general formula for $E(X^t)$ when X has the log-normal distribution Finding the answer to this integral is looking trivial to me.
I have to start with that:
$$E(X^t)=\int_0^\infty x^t \frac
{1}{\sqrt {2\pi}x}e^{-\ln(x)^2/2}dx$$
I am basically stuck on this integral: with $\ln x=y$ 
$$\frac{1}{\sqrt {2 \pi}}\int_0^\infty e^{yt} e^{-y^2/2}dy$$
I know that the answer is: $$e^{\frac{t^2}{2}}$$
 A: Hint:$$e^{yt}e^{-y^2/2}=e^{-\frac12(y^2-2yt+t^2)+\frac12t^2}=e^{t^2/2}e^{-\frac12(y-t)^2}$$
A: (I believe you've made two small mistakes - in the second integral it should be $dy$ instead of $dx$, and the integral should be taken over $-\infty$ to $\infty$. Be careful when doing substitutions!)
The trick here is in completing the square:
$$
\begin{align}
e^{yt} e^{-y^2/2} &= e^{-\frac{1}{2} (y^2 - 2yt)} \\
&=e^{-\frac{1}{2} (y^2 - 2yt+ t^2)}e^{\frac{1}{2}t^2}\\
&=e^{-\frac{1}{2} (y-t)^2}e^{\frac{1}{2}t^2}\\
\end{align}
$$
Putting this expression back into the original integral that you were stuck on, you'll get:
$$
\begin{align}
\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} (y-t)^2}e^{\frac{1}{2}t^2}dy
&=e^{\frac{1}{2}t^2}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} (y-t)^2}dy
\end{align}
$$
Observing that what you have in the integral is the PDF of the Gaussian distribution, which integrates out to 1, you will get what you want:
$$
\begin{align}
e^{\frac{1}{2}t^2}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} (y-t)^2}dy
\end{align} = e^{\frac{1}{2}t^2}
$$
