In the Handbook of Integral Equations (1.4-1, 1.4-2), solutions are given for $y$ in the equations
$$\int_0^x{\ln(x-t)y(t)}\ dt = f(x)$$
and
$$\int_0^x{(\ln x-\ln t)y(t)}\ dt = f(x)$$
I’m interested in solving
$$\int_0^x{(\ln(x-t)-\ln x)y(t)}\ dt = f(x)$$
I tried using variable substitutions or a combination of the above two problems, but couldn’t make anything work.
The equation is equivalent to \begin{equation} \int_0^x\ln\left( 1-\frac{t}{x} \right)y(t)\,dt=f(x) \end{equation} or \begin{equation} \int_0^1\ln\left( 1-u \right)y(ux)\,du=\frac{f(x)}{x} \end{equation} A solution can be found if $f(x)$ can be represented as a series \begin{equation} f(x)=x^\lambda\sum_{n=0}^N f_nx^n \end{equation} we can express \begin{equation} y(x)=x^\lambda\sum_{n=0}^N b_nx^n \end{equation} By identifying the coefficients, we have \begin{equation} y(x)=x^\lambda\sum_{n=0}^N\frac{a_n}{I_n}x^{n-1} \end{equation} where \begin{equation} I_n=\int_0^1 \ln(1-u)u^{n+\lambda-1}\,du \end{equation} we have (G\&R 4.293.8) \begin{align} I_n&=-\frac{1}{n+\lambda}\left[\psi(n+\lambda+1)-\psi(1)\right]\\ \end{align} where $\psi$ is the digamma function. If $\lambda=0$, we have $I_0=-\pi^2/6$.