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Let's consider a complex function that can be represented in the following form: $$ K(z)=\int_{-\infty}^{\infty}A(\alpha)z^\alpha d\alpha $$ 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.

The fist formula seems a generalized taylor or Laurent series. If we have a function that is well defined over $\mathbb{C}$, like a holomorphic function, we find that this formula gives the same result of the Taylor series! It vanishes when $\alpha$ is not an integer and gives a delta times its $n$-th derivative when its an integer $n$. But when we choose a function that is well defined in another riemann surface, like $z^\alpha$, 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 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? I have the impression that there is not a unique way of doing this. In a purely formal way: $$ K(re^{i\theta})=\int_{-\infty}^{\infty}A(\alpha)r^\alpha e^{i\alpha\theta} d\alpha=\int_{-\infty}^{\infty}A(\alpha)r^\alpha \sum_{n=0}^{\infty}\frac{(i\alpha\theta)^n}{n!} d\alpha= $$ $$ \sum_{n=0}^{\infty}(i\theta)^n(r\partial_r)^n\int_{-\infty}^{\infty}A(\alpha)r^\alpha d\alpha=e^{i\theta r\partial_r}K(r) $$ Can this give some hint of how to do it? What can we say about the support in $\alpha$ of this extension?

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Isn't there a problem with the fact that $z^{\alpha}$ is not uniquely defined ? –  Joel Cohen May 22 '12 at 1:29
    
Not really, as the inverse transform ask me to integrate over the unit circle an infinity of times, not really over the complexes, so this a flexibility. What is really important for me is the real positive part, and I would like to know on how many forms this decomposition can be made and where. –  juanrapha May 22 '12 at 9:49
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