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Exponential moment of the hyperbolic secant (∫R​eaxsech(πx)dx=sec(a/2),∣a∣<π)

Codex (@codex,  0) ... Analysis Real analysis Calculus Exponential function Hyperbolic function Hyperbolic secant
2026-10-20  0 By others on same topic  0 Discussions Create my own version
For real ∣a∣<π, integrate g(z)=eaz/coshπz around a rectangle of height i and width 2R. Only the simple pole i/2 lies inside, with residue eia/2/(πi). Since g(x+i)=−eiag(x), the reversed top edge adds eia times the bottom integral. The two vertical edges vanish as R→∞, using ∣a∣<π. The residue theorem gives (1+eia)J=2eia/2, proving the formula. The condition is also the exact convergence range: the two tails decay with exponents π−a and π+a.

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  1. Hyperbolic secant
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  • Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 4 / Solution

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