Homogeneous Fredholm Equation of Second Kind I'm trying to show that the eigenvalues of the following integral equation
\begin{align*}
\lambda \phi(t) = \int_{-T/2}^{T/2} dx \phi(x)e^{-\Gamma|t-x|}
\end{align*}
are given by 
\begin{align*}
\Gamma \lambda_k = \frac{2}{1+u_k^2}
\end{align*}
where $u_k$ are the solutions to the transcendental equation
\begin{align*}
\tan(\Gamma T u_k) = \frac{2u_k}{u_k^2-1}.
\end{align*}
My approach was to separate this into two integrals:
\begin{align*}
\int_{-T/2}^{T/2} dx \phi(x)e^{|t-x|} = e^{-\Gamma t}\int_{-T/2}^t dx \phi(x)e^{\Gamma x} + e^{\Gamma t} \int_t^{T/2} dx \phi(x) e^{-\Gamma x}.
\end{align*}
Then I differentiated the eigenvalue equation twice with this modification to find that
\begin{align*}
\ddot{\phi} = \frac{\Gamma^2-2\Gamma}{\lambda}\phi
\end{align*}
indicating that $\phi(t)$ is a sum of exponentials $Ae^{\kappa t} + Be^{-\kappa t}$ where $\kappa^2$ is the coefficient in the previous equation.  Can anyone confirm this is the correct approach?  I don't think there's any way to factor the original kernel and invert the result.   And I'm having trouble determining the initial conditions of the equation to set $A$ and $B$.  Assuming I can find these my instinct would be to plug $\phi$ back into the original equation and explicitly integrate and solve for $\lambda$.

This problem was solved in Mark Kac, Random Walk in the presence of absorbing barriers, Ann. Math. Statistics 16, 62-67 (1945)  https://projecteuclid.org/euclid.aoms/1177731171
