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We can expand any analytic around a point function in a Taylor series around the same point (I consider real functions for now).

$$f(x)=\sum_{k=0}^\infty f^{(k)}(x_0) \frac{(x-x_0)^k}{k!}$$

By using fractional calculus, can we also represent non-analytic functions in a kind of a 'Taylor integral'?

$$g(x)=\int_0^\infty g^{(a)}(x_0) \frac{(x-x_0)^a}{\Gamma(a+1)}da$$

Here $g^{(a)}$ is a fractional derivative.

My reasoning is as follows. We have Fourier series for periodic functions and Fourier integral (Fourier transform) for non-periodic functions. Could it work the same way for Taylor series?

There is a very similar question even with almost the same notation (I've just found it though). However, it doesn't use the concept of fractional derivative.

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    Have you thought about writing your expansion on the Fourier side, where fractional differentiation is just pointwise multiplication? – 2017-01-09
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    See whether this helps you: http://epubs.siam.org/doi/abs/10.1137/0502004 – 2017-01-09
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    And the Taylor expansion (with remainder) is obtained by [integration by parts](https://en.wikipedia.org/wiki/Taylor's_theorem#Derivation_for_the_integral_form_of_the_remainder). You should obtain some similar expressions with the fractional derivative. – 2017-01-09

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