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    • Thread Starter

    For g: (0,\infty)\to\mathbb{R} decaying sufficiently fast at \infty and \alpha\in(0,1), the \alpha-Abel transform is defined as
    \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
    \displaystyle \int_{x}^{s}\frac{dt}{(t-x)^{ \alpha}(s-t)^{1-\alpha}}=\Gamma( \alpha)\Gamma(1-\alpha)
    to show that, for sufficiently smooth g, we have ( -\partial_{x}A_{1-\alpha}\circ A_{\alpha})g=g,

    We can compute

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

    \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)=\int_{t}^{\infty}\frac{g(s)  }{(s-t)^{1-\alpha}}\,ds

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

     \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

    \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


    \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.
    • Thread Starter


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

    \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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Updated: January 28, 2017
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