Could be tetration if this integral converges

[mike3]

And the fractional derivatives can be written using the formula for the Weyl differintegral, thus an integral transform. So how do you get \( \beta(x) \) from \( F(z) \)?

\( F(z) = \frac{1}{\G(z)} \int_0^\infty \beta(x)x^{z-1}\,dx \)

\( \beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dx \)

If these integrals are absolutely convergent in the strip \( a < \Re(z) = \sigma < b \)

But I was wondering if it was possible to work in the other direction, starting with the definition for periodic functions and then expanding it to the integral-transform definition, and so if something similar could be done for the forward difference operator.

Ohhhh I see. Sorry I've never actually read many papers on the periodic definition of the weyl differintegral. I've only read papers by Erdelyi and some others and they always talk about the weyl differintegral starting from the transform that I use.

I'm certain something similar can be done to the forward difference operator. It won't look as elegant I think though.

Is there a mistake in the above integrals? In the integral for \( \beta(x) \), \( x \) appears as both function argument and as dummy variable inside the integral. Do you mean \( \beta(z) \)? But if that's so, then what's the need for the integral -- the \( \Gamma(z) \) and \( F(z) \) do not depend on the integration variable and so come out, leaving you with only the integral of the power function, which can be solved explicitly?

Or do you mean

\( \beta(z) = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \Gamma(x) F(x) x^{-z} dx \)

?

Also, what about what I mentioned earlier with regard to the tetration integral not working?