# Fourier transform of $x^\alpha$

Define $\hat{f}(\xi)\equiv F(f)(\xi):=\int_{\mathbb{R}}e^{ix.\xi}f(x)dx$

My question is: if we consider $x^{\alpha}$ as a distribution then what is $F(x^{\alpha})(\xi)$ where $0<\alpha\in\mathbb{R}$. thanks

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Praise the lord Wolfram!! wolframalpha.com/input/?i=FourierTransform+x^5 –  yohBS Feb 13 '12 at 7:16
@steven how do you define $x^{\pi}$ for $x<0$? –  Andrew Feb 13 '12 at 7:56
The distribution defined by $x^\alpha$ (supported on $\mathbb R_{>0}$) is an example of a homogeneous distribution of degree $\alpha$. It is pretty clear (if you use proper notation) that the Fourier transform of a homogeneous distribution of degree $\alpha$ on $\mathbb R^n$ is homogeneous of degree $-\alpha-n$. To get everything exactly correct requires some work. This stuff is well known and should be in most/many books on distribution theory. E.g., see Vol I of Hormander's book "The Analysis of Linear Partial Diff. Ops", Example 7.1.17. –  B R Apr 14 '12 at 1:37
According to Maple, for positive integer $k$, the Fourier transform of $x^k$ is a certain constant multiple of the $k$th derivative of the Dirac delta function. (Start with the Fourier transform of the constant 1 as $2\pi$ times the delta function, then use the rule on how multiplying by $x$ corresponds to differentiating the transform.) –  GEdgar May 15 '12 at 2:53
I doubt this function has a Fourier transform. A sufficient condition for a function $f$ to have a Fourier transform is $$||f||_{1}:=\int_{-\infty}^{\infty}|f(x)|\, \mathrm{d}x < \infty.$$ The function $f(x):=x^{\alpha}$, $\alpha \in \mathbb{R}^{+}$, does not satisfy this condition.
but if $x^\alpha\in L_{loc}(\mathbb{R})$, then it defines a distribution? my question is what is the Fourier transform of this distribution ? –  steven Feb 13 '12 at 7:05
Would you be so kind as to define $L_{\mathrm{loc}}(\mathbb{R})$ for me? –  Nick Thompson Feb 13 '12 at 7:22
Riffing off the comment above, $\text{FourierTransform}\left[x^{\alpha },x,\omega ,\text{Assumptions}\to \alpha >0\right]$ returns $\frac{e^{\frac{i \pi \alpha }{2}} \text{Abs}[\omega ]^{-1-\alpha } (-\omega +\text{Abs}[\omega ]) \text{Gamma}[1+\alpha ] \text{Sin}[\pi \alpha ]}{\sqrt{2 \pi } \omega }$. –  Nick Thompson Feb 13 '12 at 7:38
@NickThompson $L^1_{loc}(\mathbb{R})$ is defined as the class of functions which are in $L^1$ when restricted to a compact set. Any power of $x$ is well-defined as a tempered distribution if you restrict it to positive $x$ (it is important that you consider the class of tempered distributions here, since $C_0^\infty(\mathbb{R}^d)$ is not fixed by the Fourier transform, while the Schwarz class is), so it makes sense to consider the Fourier transform of such a function. –  Chris Janjigian Apr 14 '12 at 1:11
Don't you have to define $x^\alpha$ for $x<0$ and $\alpha$ irrational before you claim it is in $L_\mathrm{loc}$? –  GEdgar May 15 '12 at 2:40