We start with a little of context. I needed that a function from $\mathbb{R}^+$ to $\mathbb{R}$ could be represented in the following form, not necessarily uniquely:
$$
K(z)=\int\_{-\infty}^{\infty}A(\alpha)z^\alpha d\alpha
$$
with some hope that this big collection would give me the possibility of saying things like "my function goes to zero at infinity and is integrable in a neighborhood of 0, so I can have a decomposition with support in $(-1,0)$". Starting formally, over the complex numbers and writing $z=re^{i\theta}$, we get:
$$
K(re^{i\theta})=\int\_{-\infty}^{\infty}A(\alpha)r^\alpha e^{i\alpha\theta} d\alpha
$$
It looks like the inverse Fourier transform of $A(\alpha)r^\alpha$, then we can write:
$$
A(\alpha)=\frac{1}{2\pi}\int\_{-\infty}^{\infty}K(re^{i\theta})r^{-\alpha}e^{-i\alpha\theta}d\theta
$$
Writing it in terms of $z$ once more, we get:
$$
A(\alpha)=\frac{1}{2\pi i}\int\_C \frac{K(z)}{z^{\alpha+1}}dz
$$
where $C$ is the circuit that runs the circle of radius $r$ in both senses an infinite number of times. This is the first problem: **Why this arbitrarity in the choice of $r$**? It seems the property of a holomorphic function, but nothing is assumed here. Maybe every function with a decomposition like this is holomorphic somewhere (like $z^\alpha$ over its riemann surface)?


The fist formula seems a generalized taylor or Laurent series. If we have a function that is well defined over $\mathbb{C}$, *we find that this formula gives the same result of the Laurent series*! And we know that it converges in its domain of holomorphy (if it exists). But when we choose a function that is well defined in another riemann surface, like a function of $z^\alpha$, it gives a series in powers of $\alpha$, like a fractional Taylor series. So this definition of the inverse transform has some connections with the fractional derivative given by Nekrassov or [Osler](http://www.rowan.edu/open/depts/math/osler/LeibRuleFractDerivGen.pdfig2=uurm8yrsxD4djl4PIEcqQg) in terms of complex integration.


But my question is: **How and when can we extend a real valued function over $\mathbb{R^+}$ to a function in $\mathbb{C}$ (or some Riemann surface) that has this form?** Or some characterization of how and when to do it in some related way, as an holomorphic function for example. I have the impression that there is not a unique way of doing this, as there are many ways to "turn around the complexes" with a function in $\mathbb{R}^+$. In a purely formal try:
$$
K(re^{i\theta})=\int\_{-\infty}^{\infty}A(\alpha)r^\alpha e^{i\alpha\theta} d\alpha=\int\_{-\infty}^{\infty}A(\alpha)r^\alpha \sum\_{n=0}^{\infty}\frac{(i\alpha\theta)^n}{n!} d\alpha=
$$
$$
\sum\_{n=0}^{\infty}(i\theta)^n(r\partial\_r)^n\int\_{-\infty}^{\infty}A(\alpha)r^\alpha d\alpha=e^{i\theta r\partial\_r}K(r)
$$
And inverting:
$$
A(\alpha)=\frac{1}{2\pi}\int\_{-\infty}^{\infty}r^{-\alpha}e^{-i\alpha\theta}e^{i\theta r\partial\_r}d\theta K(r)
$$
That would mean that there is an operator that given the positive reals function would give us the coefficients $A$.


*Can this give some hint of how to do it?* What can we say about the support in $\alpha$ of this extension?


Thank you!



About an integral transform or generalized Laurent series
See the *two-sided Laplace transform*: [WIKIPEDIA](http://en.wikipedia.org/wiki/Two-sided_Laplace_transform)  

In $\int\_{-\infty}^\infty e^{-st} f(t)\;dt$ write $z=e^{-s}$.


