How to find nth moment?

I’m quite new to the field so please bare with me.

Problem: Let ξ be a random variable distributed according to a log-normal distribution with parameters μ and $σ^2$, i.e. log(ξ) is normally distributed with mean μ and variance $σ^2$. Show that the kth moment of ξ is given by $E[ξ^k]=e^{kμ+(k^2σ^2)/2}$

Question: I suppose for this problem we can use the pdf of log-normal distribution and by the definition of the nth moment we need to take an integral $\int_{-\infty}^{\infty}x^nf(x)dx$ where f(x) is our pdf. However what is our x?

P.S.: I am confused by the log(ξ), it seems that we needed to kinda switch the representation of x to log(x) and then use the pdf of normal distribution to obtain a solution for random variable y, which corresponds to y=log(x). Once we get that we can simply find x=exp(y). Or is it just a different approach and we can boldly use log-normal distribution?

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First of all observe that the moment generating function of a Gaussian random variable $X$ is:
M(t)=\mathbb{E}(e^{tX})=e^{\mu t+\frac 12\sigma^2 t^2}.
Given that $\log\xi$ is normally distributed, we can see:
\mathbb E(\xi^n)=\mathbb E(e^{n\log\xi})=M(n)=e^{\mu n+\frac 12\sigma^2 n^2}.

To explicitly answer your question, to find $n$-th moment of a random variable $\xi$, means to find $\mathbb E(\xi^n)$. Therefore if $\xi$ is a log-normal RV, the $n$-th moment is as follows:
\mathbb E(\xi^n)=\int_0^\infty \xi^n\frac{1}{\xi\sqrt{2\pi\sigma^2}}e^{-\frac{(\log\xi-\mu)^2}{2\sigma^2}}d\xi.

A “moment” of Q is the integral $\int x\; dq$. This means that you integrate the coordinate on some axis (eg radius), over the elements of Q. The n’th moment is $\int x^n \; dq$.

Volume, for example, is ‘moment of surface’. This is because surface can be treated as a normal vector by area. If you bound a surface by a loop, you can freely vary the surface, because the implication of $\int x^0 \;dq=0 $ is that the volume does not depend on the position of the observer’s coordinate. This means that for any surface bounded by a loop, the vector area is identical regardless of the shape of the surface.

Likewise, one can see that a dipole moment is simply the $p = \int x \; dq$ gives a fixed vector if the total charge is zero.

Momentum is the rate of change of moment (of “mass” = weight), since while the actual moment of mass varies by the observer, the rate of change does not depend on coordinate.