If [tex]F_Y(y)[/tex] is the cumulative distribution function for [tex]Y[/tex], then
[tex]F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)[/tex]
Then the probability density function for [tex]Y[/tex] is [tex]f_Y(y)={F_Y}'(y)[/tex]:
[tex]f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}[/tex]
The [tex]n[/tex]th moment of [tex]Y[/tex] is
[tex]E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy[/tex]
Let [tex]u=\ln y[/tex], so that [tex]\mathrm du=\frac{\mathrm dy}y[/tex] and [tex]y^n=e^{nu}[/tex]:
[tex]E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du[/tex]
Complete the square in the exponent:
[tex]nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2[/tex]
[tex]E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du[/tex]
But [tex]\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}[/tex] is exactly the PDF of a normal distribution with mean [tex]n[/tex] and variance 1; in other words, the 0th moment of a random variable [tex]U\sim N(n,1)[/tex]:
[tex]E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1[/tex]
so we end up with
[tex]E[Y^n]=e^{\frac12n^2}[/tex]
