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I am curious if there is a way to invert the integral $$f(x)=\int_x^a \frac{g(t)}{\sqrt{t-x}}dt$$ to solve for g(x) when f(x) is a known function. The integral from x to a makes this problem seem a little awkward.

I have found that it is possible to invert equations of the form $$ f(x)=\int_0^x \frac{g(t)}{(x-t)^\alpha}dt$$ for $0<\alpha<1$ by using the abel transform, but I dont think this applies.

Anode
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Define $L_\alpha$ by $$ (L_\alpha g)(x)=\int_0^x \frac{g(t)}{(x-t)^\alpha}\mathrm dt. $$ Then, using the change of variable $s=a-t$, one gets $$f(x)=\int_x^a \frac{g(t)}{\sqrt{t-x}}\mathrm dt=\int_0^{a-x} \frac{g(a-s)}{\sqrt{a-x-s}}\mathrm ds=(L_{1/2}h)(a-x),$$ where the function $h$ is defined by $h(t)=g(a-t)$ for every $t$. Thus, if one knows $f$ such that $$f(x)=\int_x^a \frac{g(t)}{\sqrt{t-x}}\mathrm dt$$ and if one can invert $L_{1/2}$, then one knows $h$ such that $(L_{1/2}h)(x)=f(a-x)$ hence one knows $g$ since $g(t)=h(a-t)$ for every $t$.

Did
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  • Sorry this is probably a stupid question but how did you get 0 as the lower bound of the integral after the change of variables? – Anode May 25 '14 at 17:00
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    If $x\lt t\lt a$ and $s=a-t$ then $$ $\lt s\lt$ $$. – Did May 25 '14 at 17:02