Ratio of integrals & Golden-ratio.

Note that $\begin{aligned} F(a) & = \; \int_{-\infty}^\infty \frac{1}{2\cosh x - a}\,dx \; = \; \int_{-\infty}^\infty \frac{e^x\,dx}{e^{2x} - ae^x + 1} \\ &= \; \Big[\frac{2}{\sqrt{4-a^2}}\tan^{-1}\left(\frac{2e^x-a}{\sqrt{4-a^2}}\right)\Big]_{-\infty}^\infty \; = \; \frac{1}{\sqrt{4-a^2}}\left\{ \pi + 2\tan^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)\right\} \end{aligned}$ for any $|a| < 2$ , and hence $\int_{-\infty}^\infty \frac{1}{(2\cosh x - a)(2\cosh x - a^{-1})}\,dx \; = \; \frac{1}{a-a^{-1}}\big(F(a) - F(a^{-1})\big)$ for any $\tfrac12 < |a| < 2$ . The ratio we are interested in is $\alpha \; = \; \frac{F(\phi) - F(\phi^{-1})}{F(-\phi^{-1}) - F(-\phi)}$ (recall that $\phi - \phi^{-1} = 1$ ). After much simplification, we obtain $\alpha \; = \; \tfrac12(7 + 5\sqrt{5})$ and hence $\alpha^2 - 7\alpha = \boxed{19}$ .