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I need to find the resolvent kernel of $\phi(x)=f(x)+\int\limits_0^x \sin(x-t)\phi(t)\,dt$.

Differentiating both sides I get $$ \phi'(x) = f'(x) + \int_0^x \cos(x-t)\phi(t)\,dt. $$

And now I am stuck.

I have also tried the resolvent kernel sum method, ie. \begin{gather*} K(x,t) = K_1(x,t) = \sin(x-t) \\ K_2(x,t) = \int_0^x K(x,s)\cdot K(s,t)\,ds = x\cos(x-t)-\frac{1}{2}\sin(t)\cos(x) \end{gather*} and so on...

Stefan Hante
  • 2,616
Dharmaraj Deka
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  • No ! You are not allowed to differentiate in this way: the fact that variable $x$ is present also at one of the bounds of integration necessitates that you use a more general formula that you can find for example in the answer to the following question (https://math.stackexchange.com/q/1223409). – Jean Marie May 18 '17 at 11:10
  • try using $\phi''(x) = f''(x) + f(x)$ – Manglu May 18 '17 at 11:15

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