Question

Abel's integral equation which is f(x)=\int_{x}^{\infty}\frac{dtg(t)}{(t-x)^\alpha } where (0<\alpha <1) and g(t)=-\frac{sin\pi\alpha}{\pi}\frac{d}{dt}\int_{t}^{\infty}\frac{dxf(x)}{(x-t)^{1-\alpha}}=-\frac{sin\pi\alpha}{\pi}\int_{t}^{\infty}\frac{dx}{(x-t)^{1-\alpha}}\frac{df}{dx} . invert this j(r)=2\int_{r}^{\infty}dr\frac{rj(r)}{\sqrt{r^2-r^2}} using the equations above to get j(r)=-\frac{1}{\pi}\int_{r}^{\infty}\frac{dr}{\sqrt{(r^2-r^2)}}\frac{di}{dr}

Answer 1

$Abel's \: integral \: equation \: which \: is \\ f(x)=\int_{x}^{\infty}\frac{dtg(t)}{(t-x)^\alpha } \\ where (0<\alpha <1) \\ and \\ g(t)=-\frac{sin\pi\alpha}{\pi}\frac{d}{dt}\int_{t}^{\infty}\frac{dxf(x)}{(x-t)^{1-\alpha}}=-\frac{sin\pi\alpha}{\pi}\int_{t}^{\infty}\frac{dx}{(x-t)^{1-\alpha}}\frac{df}{dx} . \\ invert \: this \\ j(r)=2\int_{r}^{\infty}dr\frac{rj(r)}{\sqrt{r^2-r^2}} \\ using \: the \: equations \: above \: to \: get \\ j(r)=-\frac{1}{\pi}\int_{r}^{\infty}\frac{dr}{\sqrt{(r^2-r^2)}}\frac{di}{dr}$
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