# Fourier Transform of $g(x) = (1 + x^{1/a} )^a$

Given a function

$g(x) = (1 + x^{1/a} )^a \times 1_{\{x > 0\}}$

that is bounded that is bounded by zero, can I compute the fourier transform?

Thanks!

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The Fourier integral does not converge for any real value of $a$. – Ron Gordon Sep 15 '13 at 14:42
Because its not 'nice' in the sense of decaying at infinity. Are there values of $a$ for which it is possible to obtain the generalised fourier transform? – Luap Nalehw Sep 15 '13 at 18:45

For a generalized Fourier transform when $a$ is an integer, expand $(1 + u)^a$ via the binomial theorem and set $u = x^{\frac{1}{a}}$. You'll get

$\sum_{k=0}^a\binom{a}{k}x^{\frac{k}{a}}$.

Then apply the Fourier transform identity:

$x^{\frac{k}{a}}f(x) \xrightarrow{\mathscr{F}} (\frac{i}{2\pi})^{\frac{k}{a}}\frac{d^\frac{k}{a}}{d\zeta^\frac{k}{a}}\mathscr{F}f(x)$

to the series term by term, with $\mathscr{F}f(x) = \delta(\zeta)$.

For the fractional derivatives of the delta function, you can apply the identity:

$D^\alpha\delta(\zeta) = \dfrac{1}{\Gamma(\lceil\alpha\rceil-\alpha)}\dfrac{d^{\lceil\alpha\rceil}}{d\zeta^{\lceil\alpha\rceil}}(\zeta^{\lceil\alpha\rceil-\alpha-1}H(\zeta))$.

I think the same idea could be applied with $a$ real, but the series would be infinite.

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I solved this problem via numerical inversion in the end but would like to follow up on your idea. Quick question - where does this fourier identify come from? Thanks – Luap Nalehw Oct 22 '14 at 9:50