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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 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!

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1 Answer 1

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See the two-sided Laplace transform: WIKIPEDIA
In $\int_{-\infty}^\infty e^{-st} f(t)\\,dt$ write $z=e^{-s}$.

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Thank you for your answer! I had already searched a little about that, but it is the inverse that bugs me. Its like if we had the Laplace transform of a function over the positive reals, but we need it over the complex plane to calculate it with Mellin's inverse formula, so the problem just changes of name. –  guaraqe May 22 '12 at 17:00
    
Finally, it helped me very much (after some research). Thank you! –  guaraqe May 31 '12 at 21:10

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