In the study of few-particle quantum systems, I have come upon the following integral equation: \begin{equation} f(x) = \lim_{\Lambda\rightarrow \infty}\frac{3\sqrt{3}}{\ln(4x^2+6\lambda) + 2\gamma} \int^{\Lambda}_{-\Lambda}dy\ \frac{f(y)}{\sqrt{(x+y)^2+2x^2+2y^2+4\lambda}}, \end{equation} where $\lambda$ is real and positive; $\gamma$ is the Euler-Mascheroni constant. Though the denominator diverges logarithmically, I anticipate that $f(x)$ falls off sufficiently quickly/oscillates such that the integral converges. In any case, the first-order result in $1/\Lambda$ is acceptable as well. I have attempted series expansions, trig substitutions, and guess-and-check functions to no avail. I considered trying to find a differential operator that would satisfy the kernel as a Green's function, but I have never encountered a Green's function that was not symmetric. So far, numerical attempts (varying $\lambda$) have not converged.
Any tips towards a solution or an approach would be greatly appreciated.
Since the integration bounds will be taken to infinity, we can shift the integration variable to rewrite the expression as \begin{equation} f(x)\left[\ln\left(4x^2 + 6\lambda\right) + 2\gamma\right] = 3 \int dy\ \frac{f\left(y-\frac{x}{3}\right)}{\sqrt{y^2+\frac{2}{9}\left(4x^2+6\lambda\right)}}, \end{equation} where the denominator is seen to be the square root of a Breit-Wigner distribution (a.k.a. Lorentzian, Cauchy). Consider now the integral in the complex plane, \begin{equation} \int dz\ \frac{f\left(z-\frac{x}{3}\right)}{\left(z-z_0\right)^{1/2} \left(z+z_0\right)^{1/2}}, \end{equation} where $z_0 = i \frac{\sqrt{2}}{3} \sqrt{4x^2+6\lambda}$. With the $1/2$ powers, we have branch points rather than poles, so we choose a contour in the upper half plane that avoids the branch point at $z=z_0$, as in Contour integral with branch point. With no poles enclosed, the contour integral will evaluate to zero. The contours from $z=0\rightarrow z=z_0$ and back cancel in this case; on physical grounds, we posit that $|f(z)|\rightarrow 0$ as $|z|\rightarrow 0$, giving zero for the arc at infinity. The result of all this is that the original integral of interest is equal to the circular integral around the branch point.
Making the variable change $z-z_0 = \rho e^{i\theta}$, that integral can be written as \begin{equation} \lim_{\rho\rightarrow 0} \int_{-\pi/2}^{3\pi/2} d\theta\ i\rho e^{i\theta} \frac{f\left(\rho e^{i\theta} +z_0 - \frac{x}{3}\right)}{\left(\rho e^{i\theta}\right)^{1/2} \left(\rho e^{i\theta}+2z_0\right)^{1/2}}. \end{equation} Because of the $\rho^{1/2}$ appearing in the integrand, the integral will vanish unless \begin{equation} f(z) = \frac{1}{\sqrt{z-z_0 +x/3}} \sum_{k=0}^{\infty} \frac{a_k}{\left(z-z_0+x/3\right)^k}, \end{equation} where the sum arises from expanding $(\rho e^{i\theta} +2z_0)^{1/2}$ as a binomial series; we consider only the $k=0$ term now for clarity. In that case, \begin{equation} \lim_{\rho\rightarrow 0} \int_{-\pi/2}^{3\pi/2} d\theta\ i\rho^{1/2} e^{i\theta/2} \frac{f\left(\rho e^{i\theta} +z_0 - \frac{x}{3}\right)}{\left(\rho e^{i\theta}+2z_0\right)^{1/2}} = \frac{2 \pi i a_0}{\sqrt{2z_0}}. \end{equation} Applying the solution of $f$ to both sides (and including all powers of $k$), we obtain \begin{equation} \frac{\ln\left(4x^2 + 6\lambda\right) + 2\gamma}{\sqrt{\frac{4}{3}x-z_0}} \sum_{k=0}^{\infty} \frac{a_k}{\left(\frac{4}{3}x-z_0\right)^k} = \frac{6\pi i}{\sqrt{2z_0}} \sum_{k=0}^{\infty} \binom{-1/2}{k} \frac{a_k}{\left(2z_0\right)^k}. \end{equation} As verified by the Wronskian for two terms of different powers, each term is linearly independent as functions of $x$, so that the equation holds term by term: \begin{equation} \frac{\ln\left(4x^2 + 6\lambda\right) + 2\gamma}{\left(\frac{4}{3}x-z_0\right)^{k+1/2}} = \frac{6\pi i}{\left(2z_0\right)^{k+1/2}} \binom{-1/2}{k},\ k=0,1,2,... \end{equation} Finding the eigenvalues $\lambda$ that fulfill this condition is another matter, but this at least provides the functional form.