I'm looking for $I=\int_{0}^\infty f(u)du$ where $f$ is the solution to the integral equation $$\frac{\pi}{2}f(u) = \frac{1}{1+u^2} - \int_{0}^\infty \frac{f(v) dv}{4+(u-v)^2}$$ for $u$ between $0$ and $\infty$.
While knowing $f(u)$ exactly would be nice, I'd be quite happy with just $I=\int_{0}^\infty f(u)du$ or an approximation to $I$! My goal would be an approximation within $1$%.
I have two attempts below - one with an ansatz for $f(u)$ and the other a perturbative series for $I$. I'm hopeful that either direction will be fruitful.
Here's a sketch of my attempt that roughly estimates $I \approx .6$.
First, note that a problem with different bounds and with $u$ between $\infty$ and $\infty$, $$\frac{\pi}{2}g(u) = \frac{1}{1+u^2} - \int_{\color{red}{-\infty}}^\infty \frac{g(v) dv}{4+(u-v)^2}$$ can be solved via Fourier transform to find $g(u) = \frac{1}{e^{\frac{\pi}{2}u}+e^{-\frac{\pi}{2}u}}$ and $\int_{0}^\infty g(u) du = \frac{1}{2}$.
I anticipate that $I$ for the problem of interest will be a little more than $.5$, given that it appears we're subtracting less from the right hand side when we use smaller bounds.
Consider the ansatz $$f(u) = \frac{c_1}{e^{\frac{c_2 \pi}{2}u}+e^{-\frac{c_2\pi}{2}u}}$$ Changing the parameters $c_1$ and $c_2$ by hand, I find with numerical integration that $c_1=1.11$ and $c_2=.93$ make the left hand side and right hand side of $$\frac{\pi}{2}f(u) = \frac{1}{1+u^2} - \int_{0}^\infty \frac{f(v) dv}{4+(u-v)^2}$$ nearly equal as a function of $u$. I anticipate this ansatz won't include the exact solution, but it gives what I think is a reasonable estimate for $I$ of about $.6$.
Here is another attempt. Note that these linear problems can in principle be solved through perturbative solutions; $$f(u) = h(u)+\lambda \int_0^\infty K(u,v) f(v)$$ is solved by $$f(u) = \sum_{n=0}^\infty \lambda^n \int_0^\infty du_2 ... \int_0^\infty du_{n+1} K(u, u_2)...K(u_n, u_{n+1}) h(u_{n+1})$$ so $$I = \sum_{n=0}^\infty \lambda^n \int_0^\infty du_1 \int_0^\infty du_2 ... \int_0^\infty du_{n+1} K(u_1, u_2)...K(u_n, u_{n+1}) h(u_{n+1}).$$
That is, we have $$I = \sum_{n=0}^\infty \left(-\frac{2}{\pi}\right)^n \frac{2}{\pi} \int_0^\infty du_1 ... \int_0^\infty du_{n+1} \frac{1}{4+(u_1-u_2)^2}...\frac{1}{4+(u_n-u_{n+1})^2}\frac{1}{1+u_{n+1}^2}$$
The first few partial sums are $1, 0.318, 0.866$. Assuming the terms are decreasing in the series, the alternating sum is then bounded between $0.318<I<0.866$. This is consistent with $I\approx.6$, but the bounds are quite large, and the integrals rapidly become both analytically and computationally challenging to evaluate.