We were asked to show that if the following integral converges:
$$ \mu_k =\int_{-\infty}^{\infty} x^k \space f(x)dx \space, k \in \mathbb{N}$$
Then we can obtain $ \mu_k $ from the Fourier Transform of $f(x)$ without the need for direct integration - I'm stumped for this one - would using a Convolution be of any use here? Any gentle hints would help greatly.
Edit: The transform convention I was asked to use was (if it matters in any way):
$$ F(q) = \int_{-\infty}^{\infty}e^{iqx}f(x) dx$$
We have $$F\left( q \right) = \int_{ - \infty }^\infty {f\left( x \right){e^{iqx}}dx} = \int_{ - \infty }^\infty {f\left( x \right)\sum\limits_{k = 0}^\infty {\frac{{{{\left( {iqx} \right)}^k}}}{{k!}}} dx} = \sum\limits_{k = 0}^\infty {\frac{{{i^k}{q^k}}}{{k!}}\int_{ - \infty }^\infty {{x^k}f\left( x \right)dx} } = \sum\limits_{k = 0}^\infty {\frac{{{i^k}{q^k}}}{{k!}}{\mu _k}} = \sum\limits_{k = 0}^\infty {\frac{{{i^k}{\mu _k}}} {{k!}}{q^k}} $$ On the other hand, using the Taylor expansion of $F(q)$ around $q=0$, we obtain $$F\left( q \right) = \sum\limits_{k = 0}^\infty {\frac{{{F^{\left( k \right)}}\left( 0 \right)}}{{k!}}{q^k}}.$$ Thus $$\frac{{{i^k}{\mu _k}}}{{k!}} = \frac{{{F^{\left( k \right)}}\left( 0 \right)}} {{k!}}$$ for all $k=0,1,2,\ldots$, which implies $${\mu _k} = \frac{{{F^{\left( k \right)}}\left( 0 \right)}}{{{i^k}}},\,\, k=0,1,2,\ldots.$$ That means we can compute the moments $\mu_k$ via the Fourier transform $F(q)$.
