Why is $-\int_{-\infty}^\infty \log\left[1+2f'(x)(1-\cos\phi)\right]\,dx$ equal to $\phi^2$?

It is clear that $I(0)=0$, hence by Leibniz's integration rule, it suffices to show that $$\int_{-\infty}^\infty\frac{-2f'(x)\sin\phi}{1+2f'(x)(1-\cos\phi)}\,dx=2\phi,\qquad |\phi|<\pi.$$ We can assume, without loss of generality, that $0<\phi<\pi$. By a bit of algebra, we can rewrite the last equation as $$\int_{-\infty}^\infty\frac{e^x\sin\phi}{e^{2x}+2(\cos\phi)e^x+1}\,dx=\phi.$$ By the change of variables $u=(e^x+\cos\phi)/\sin\phi$, this becomes $$\int_{\frac{\cos\phi}{\sin\phi}}^\infty\frac{du}{u^2+1}=\phi,$$ which in turn can be verified by calculus: $$\int_{\frac{\cos\phi}{\sin\phi}}^\infty\frac{du}{u^2+1}=\arctan(\infty)-\arctan\left(\frac{\cos\phi}{\sin\phi}\right)=\frac{\pi}{2}-\left(\frac{\pi}{2}-\phi\right)=\phi.$$ The proof is complete.

We have $$I(t)=-\int_{-\infty}^\infty l_x(t)\,dx,$$ where $$l_x(t):=l(t):=\ln(1+2f'(x)(1-\cos t)).$$ Next, $$l_x''(t)=l''(t)=-\frac{2 e^x \left(c \left(e^{2 x}+1\right)+2 e^x\right)}{\left(2 c e^x+e^{2 x}+1\right)^2},$$ where $c:=\cos t\in(-1,1]$ for $|t|<\pi$. So, by substitution $e^x=u$, for $c\in(-1,1]$, $$I''(t)=-\int_{-\infty}^\infty l''_x(t)\,dx=2,$$ which implies $I(t)=t^2$ for $|t|<\pi$ (since $l_x(0)=0=l'_x(0)$ and hence $I(0)=0=I'(0)$).