I'm trying to solve this integral equation to find $y(x)$ but am struggling. Note, $a$ and $c$ are just two parameters.
$$\int_0^{\infty}y(t)\,\text{ln}\left|{\frac{t-x}{t+x}}\right|dt=\pi\left[\pi+2\text{tan}^{-1}(x)-\text{tan}^{-1}\left(\frac{x}{a}\right)-\text{tan}^{-1}\left(\frac{x}{c}\right)\right]$$
Any suggestions would be greatly appreciated!
Ok, so, the way I found to the solution is the following.
First, I asked myself, what would be a differential problem related to that integral equation? You know, if you have a differential problem:
$$ D f(x) = y(x) $$ where $ D $ is a differential operator,
and if you have a Green's function such that $ D G(x,t) = \delta(x-t) $, then
$$ f(x) = \int G(x,t) y(t) $$
is a solution. You can make the identification
$$ f(x) = \pi \left[ \pi + 2 \tan^{-1}(x) - \tan^{-1} \left(\frac{x}{a}\right) - \tan^{-1} \left(\frac{x}{c}\right) \right] $$
and
$$ G(x,t) = \ln \left| \frac{ t - x }{ t + x } \right| $$
and then the difficult question is: what is the differential operator $ D $?
Rather than obtaining it from calculation, I answered that question using some physics. The Green's function for the Laplacian in two dimensions is a logarithm. More precisely, consider the problem:
$$ \nabla^2 g(x) = \delta(0) $$
By the spherical simmetry of the problem, I know $ g $ depends only on $ r $, the radial distance from origin. Then I need to write the laplacian in polar coordinates, which in two dimensions read:
$$ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial g(r)}{\partial r} \right) $$
For that to become 0 outside of the origin, I need the term inside the parenthesis to be a constant, what is easily obtained if $ g(r) = k ln r $, k being a constant. Then, I need the integral of this expression over a circle surrounding the origin to be one, as to make the laplacian equivalent to a delta. The integral over the circle of area A reads, using Gauss's theorem in two dimensions:
$$ \int_A dA \nabla^2 (k \ln r) = \int_0^{2\pi} r d\theta \nabla(k \ln r) = 2\pi k $$
So that
$$ k = \frac{1}{2\pi} $$
And the Green's function is:
$$ g(r) = \frac{1}{2\pi r} $$
Now, let's consider the problem:
$$ \frac{1}{t} \frac{\partial}{\partial t} \left(t \frac{\partial}{\partial t} G(x,t) \right) = \delta(x-t) $$
Subject to the boundary condition
$$ G(x,0) = 0 $$
This can be seen as the physical problem of a point charge in the point (x,0) in the plane generating an electrical potential $ \phi_x(t,u) $ which vanishes at the surface t=0 (because, say, of a plane conductor being placed there), and we want to know the value $ G(t,x) = \phi_x(t,0) $ of the potential in the point (t,0). That is, the field lines extend through the whole plane, but in your problem we only care about the line u=0. That is, we have a one-dimensional problem, but with the laplacian that we want.
The electrical potential satisfies the equation:
$$ \nabla^2 \phi_x(t,u) = \delta((t,u) - (x,0)) $$
Or, using $ r = \sqrt{(t-x)^2 + u^2} $
$$ \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} \phi_x(r) \right)= \delta(0) $$
With the aforementioned boundary condition.
The standard way to solve this problem is through the method of images: you consider that there is no boundary condition, and instead there are TWO charges, one at $ (x,0) $ and one at $ (-x,0) $.
Then, the potential generated by the two is the sum of the potential which would be generated by each one:
$$ \phi = \frac{1}{2\pi} \ln r + \frac{1}{2\pi} \ln r' $$
r being the distance between the evaluation point and the first charge, and r' being the distance to the second charge.
Well, if we consider the evaluation point as being $ (t,0) $, then $ r = |t-x| $ and $ r' = |t+x| $, and
$$ G(t,x) = \frac{1}{2\pi} \ln \left| \frac{t-x}{t+x} \right| $$
Finally, we found out that this Green's function is the solution of the differential problem stated above, with that boundary condition.
This means that the $ y(t) $ that we want is given by
$$ \frac{1}{t} \frac{\partial}{\partial t} \left(t \frac{\partial}{\partial t} f(t) \right) = 2 \pi y(t) $$
Putting the expression for f, we have then
$$ y(t) = \frac{1}{2t} \frac{\partial}{\partial t} \left(t \left[ \frac{2}{1 + t^2} - \frac{a}{a^2 + t^2} - \frac{c}{c^2 + t^2} \right] \right) $$
Which can then be calculated explicitly.
Obs: I am thinking whether the fact that $ f(t) $ does not satisfy the boundary condition is a problem or not. In the differential problem it just amounts to an integration constant, or a change of ground potential. But I don't know if it affects the integral problem in an important way.
Obs2: Reading the comment from Mark Viola, it seems to me that this $ \pi^2 $ really should not be in the integral problem at all.