Fredholm Integral Equation for Bessel Function

Question

Show (or at least verify) that the solution to the integral equation $$u(x)=\frac{2}{\pi } \int_0^{\infty } \frac{\sin (x+y) u(y)}{x+y} \, dy$$ is given by $u(x)=J_0(x)$

Answer 1

We will show how to convert the integral equation into a convolution type and solve it by Fourier transforms.

First note that $\int_{-\infty }^0 \frac{\sin (x+t) J_0(t)}{x+t} \, dt=\int_0^{\infty } \frac{\sin (x+t) J_0(t)}{x+t} \, dt$. Thus the original problem is equivalent to showing that $$J_0(x)=\frac{1}{\pi } \int_{-\infty }^{\infty } \frac{\sin (x+t) J_0(t)}{x+t} \, dt$$
Since $J_0(-x)=J_0(x),$ the above equation is equivalent to $$J_0(x)=\frac{1}{\pi } \int_{-\infty }^{\infty } \frac{\sin (x-t) J_0(t)}{x-t} \, dt$$ The result follows after taking the Fourier transform of both sides and recalling the result for the Fourier transform of a convolution.

Note:$\mathscr{F}\{J_0(x\}=\frac{\sqrt{\frac{2}{\pi }} (\theta (k+1)-\theta (k-1))}{\sqrt{1-k^2}}$ and $\mathscr{F}\left\{\frac{\sin (x)}{x}\right\}=\sqrt{\frac{\pi }{2}} \theta (1-| k| )$

where $\theta (k)= \begin{array}{cc} \{ & \begin{array}{cc} 0 & k<0 \\ 1 & k>0 \\ \end{array} \\ \end{array}$