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 $\lambda$.