On the positive real axis, whose path does not cross the left-displaced cut,
$$
F_B(x)=\int_0^x\frac{dt}{1+t^2}=\arctan x
=\frac\pi2+O(x^{-1}).
$$
Since $F_B'(z)=1/(1+z^2)=O(|z|^{-2})$ near infinity, integrating this <derivative> along large arcs extends the same expansion uniformly there:
$$
F_B(z)=\frac\pi2+O(|z|^{-1}).
$$
Therefore
$$
\oint_{\gamma_R}\frac{F_B(t)}t\,dt
=\frac\pi2\oint_{\gamma_R}\frac{dt}{t}+O(R^{-1})
=i\pi^2+O(R^{-1}),
$$
and hence
$$
\boxed{\lim_{R\to\infty}\oint_{\gamma_R}\frac{F_B(t)}t\,dt=i\pi^2.}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 7E / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/7e/c/solution