Show that under certain circumstances, the integral equation $$\int_0^t (t^2-u^2)^{\alpha-1} \chi (u) du = \phi (t)$$
Possesses the solution:
$$\chi (t) = \frac{2}{\pi} \sin (\alpha \pi) \Big[ \phi (0) t^{1-2\alpha} + t \int_0^t (t^2 - u^2)^{-\alpha} \phi' (u) du \Big] $$
And determine the circumstances under which this solution will exist.
Honestly this question kind of blew me away, it was the last question of the section for a book I was working in and was significantly harder than the questions before it. For those curious, the book is Erdelyi's "Operational Calculus and Generalized Functions."
The coefficient in front of the brackets immediately reminded me of the reflection formula, so I suspect that somewhere we must get a coefficient $$\frac{1}{\Gamma (\alpha) \Gamma (1-\alpha)} = \frac{\sin(\alpha \pi)}{\pi}$$
Also I suspect that the circumstances under which this solution will exist are when $\alpha \in(0,1)$. My issue is deriving the formula. All of the problems before this were convolution-type integral equations and I am intended to use operational techniques to solve these, but I do not know how to tackle this problem with operational techniques. The $(t^2 - u^2)^{\alpha -1}$ is almost a convolution, and I tried factoring into the form $(t-u)^{\alpha -1} (t+u)^{\alpha -1}$ as well but it didn't seem to help. Any help is appreciated!
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\begin{align} \phi\pars{t} & = \int_{0}^{t}\pars{t^{2} - u^{2}}^{\alpha - 1}\,\,\,\chi\pars{u}\,\dd u = -\,\half\int_{u\ =\ 0}^{u\ =\ t}{\chi\pars{u} \over u}\, \dd\bracks{\pars{t^{2} - u^{2}}^{\alpha}} \\[3mm] & = \,\mathrm{f}\pars{t} + \half\int_{0}^{t}\pars{t^{2} - u^{2}}^{\alpha}\bracks{\chi\pars{u} \over u}' \,\dd u \\[3mm] & \mbox{where}\ \,\mathrm{f}\pars{t} \equiv -\,\half\lim_{u \to t}\bracks{% {\pars{t^{2} - u^{2}}^{\alpha} \over u}\,\chi\pars{u}} + \half\,\chi'\pars{0}t^{2\alpha}\quad\mbox{with}\,\,\,\chi\pars{0} = 0\tag{1} \\[8mm] \phi'\pars{t} & = \,\mathrm{f}'\pars{t} + \alpha\, t \int_{0}^{t}\pars{t^{2} - u^{2}}^{\alpha - 1}\,\, \bracks{\chi\pars{u} \over u}'\,\dd u \end{align}
\begin{align} &\int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\phi'\pars{v}\,\dd v \\[3mm] = &\ \int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\bracks{% \,\mathrm{f}'\pars{v} + \alpha\, v \int_{0}^{v}\pars{v^{2} - u^{2}}^{\alpha - 1}\,\, \bracks{\chi\pars{u} \over u}'\,\dd u}\,\dd v \\[3mm] = &\ \int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\mathrm{f}'\pars{v}\,\dd v + \alpha\int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\,v \int_{0}^{v}\pars{v^{2} - u^{2}}^{\alpha - 1}\,\, \bracks{\chi\pars{u} \over u}'\,\dd u\,\dd v \\[3mm] &\ \int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\mathrm{f}'\pars{v}\,\dd v + \alpha\int_{0}^{t}\bracks{\chi\pars{u} \over u}'\ \underbrace{% \int_{u}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\pars{v^{2} - u^{2}}^{\alpha - 1}\,\, v\,\dd v}_{\ds{\half\,\pi\csc\pars{\pi\alpha}}}\ \,\dd u \end{align}
$$ \int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\phi'\pars{v}\,\dd v = \int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\mathrm{f}'\pars{v}\,\dd v + \half\,\pi\alpha\csc\pars{\pi\alpha} \bracks{{\chi\pars{t} \over t} - \chi'\pars{0}} $$ In adopting the 'simple case' $\pars{~\mbox{see}\ \pars{1}~}$ $\ds{\,\mathrm{f}\pars{t} = \half\,\chi'\pars{0}t^{2\alpha}}$ we find $$ \int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\mathrm{f}'\pars{v}\,\dd v =\half\,\chi'\pars{0}\pi\alpha\csc\pars{\pi\alpha} $$ such that $$ \chi\pars{t} = 2\,{\sin\pars{\pi\alpha} \over \pi\alpha}\, t\int_{0}^{t}\pars{t^{2} - v^{2}}^{-\alpha}\,\phi'\pars{v}\,\dd v $$