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Can someone PLEASE help me compute the Abel transform!

For g:(0,∞)→Rg: (0,\infty)\to\mathbb{R} decaying sufficiently fast at ∞\infty and α∈(0,1)\alpha\in(0,1), the α\alpha-Abel transform is defined as
Aαg(t)=1Γ(α)∫t∞g(s)(s−t)1−α ds.\displaystyle A_{\alpha}g(t)=\frac{1}{\Gamma( \alpha)}\int_{t}^{\infty}\frac{g(s)}{(s-t)^{1-\alpha}}\,ds.

I want to use the formula
∫xsdt(t−x)α(s−t)1−α=Γ(α)Γ(1−α)\displaystyle \int_{x}^{s}\frac{dt}{(t-x)^{ \alpha}(s-t)^{1-\alpha}}=\Gamma( \alpha)\Gamma(1-\alpha)
to show that, for sufficiently smooth gg, we have (−∂xA1−α∘Aα)g=g,( -\partial_{x}A_{1-\alpha}\circ A_{\alpha})g=g,

We can compute
Unparseable latex formula:

[br](-\partial_{x} A_{1-\alpha}\circ A_{\alpha})g&=-\partial_{x} A_{1-\alpha}(A_{\alpha}g(t))



=−1Γ(α)∂xA1−α(∫t∞g(s)(s−t)1−α ds)\displaystyle =-\frac{1}{\Gamma( \alpha)}\partial_{x} A_{1-\alpha} \left(\int_{t}^{\infty}\frac{g(s)}{(s-t)^{1-\alpha}}\,ds\right)

Now this bit is where it gets very complicated for me. I don't know if I should take

G(t)=∫t∞g(s)(s−t)1−α dsG(t)=\int_{t}^{\infty}\frac{g(s)}{(s-t)^{1-\alpha}}\,ds

And then apply A1−αA_{1-\alpha} to that; because it gets very messy and doesn't appear to yield what I want. For example, I computed

A1−αG(t)=1Γ(1−α)∫t∞G(u)(u−t)−α du=1Γ(1−α)∫t∞∫u∞g(s)(s−u)1−α ds(u−t)−α du \displaystyle A_{1-\alpha} G(t)=\frac{1}{\Gamma(1-\alpha)}\int_{t}^{\infty}\frac{G(u)}{(u-t)^{-\alpha}}\,du=\frac{1}{\Gamma(1-\alpha)} \int_{t}^{\infty}\frac{\int_{u}^{ \infty}\frac{g(s)}{(s-u)^{1-\alpha}}\,ds}{(u-t)^{-\alpha}}\,du

=1Γ(1−α)∫t∞(∫u∞g(s)(u−t)α(s−u)1−α ds) du\displaystyle =\frac{1}{ \Gamma(1- \alpha) }\int_{t}^{\infty}\left( \int_{u}^{\infty}\frac{g(s)}{(u-t)^{ \alpha}(s-u)^{1-\alpha}}\,ds \right)\,du

Therefore

(−∂xA1−α∘Aα)g=−1Γ(α)Γ(1−α)∂x∫t∞(∫u∞g(s)(u−t)α(s−u)1−α ds) du\displaystyle (-\partial_{x} A_{1-\alpha} \circ A_{\alpha} )g= -\frac{1}{\Gamma(\alpha) \Gamma(1-\alpha)} \partial_{x} \int_{t}^{\infty}\left( \int_{u}^{\infty}\frac{g(s)}{(u-t)^{ \alpha}(s-u)^{1-\alpha}}\,ds \right)\,du

But I think that this is wrong...and in any case, it's driving me crazy! I would really appreciate some help here. :tongue:
(edited 7 years ago)
BUMP!

I was thinking of then switching the integration limits to remove the minus sign, so that we have:

1Γ(α)Γ(1−α)∂x∫−∞t∫u∞g(s)(u−t)α(s−u)1−α ds du\displaystyle \frac{1}{ \Gamma( \alpha) \Gamma(1- \alpha)} \partial_{x} \int_{-\infty}^{t} \int_{u}^{\infty} \frac{g(s)}{(u-t)^{ \alpha}(s-u)^{1-\alpha} }\,ds\,du,

but I am still stuck.

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