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

On the positive real axis, whose path does not cross the left-displaced cut,

F_{B} (x) = \int_{0}^{x} \frac{d t}{1 + t ^{2}} = arctan x = \frac{π}{2} + 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{π}{2} + O (∣ z ∣^{- 1}) .

Therefore

\oint_{γ_{R}} \frac{F _{B} ( t )}{t} d t = \frac{π}{2} \oint_{γ_{R}} \frac{d t}{t} + O (R^{- 1}) = i π^{2} + O (R^{- 1}),

and hence

R \to \infty lim \oint_{γ_{R}} \frac{F _{B} ( t )}{t} d t = i π^{2} .

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