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 Incoming links: Schläfli contour integral for Legendre polynomials

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 30D / b / i / Solution Created 2026-09-24 Updated 2026-09-25

Factor the integrand in the Schläfli contour integral for Legendre polynomials as

\frac{( z ^{2} - 1 ) ^{n}}{( z - t ) ^{n + 1}} = \frac{1}{z - t} (\frac{z ^{2} - 1}{z - t})^{n} .

Thus one may take

f (z, t) = \frac{1}{z - t}, ϕ (z, t) = lo g (\frac{z ^{2} - 1}{z - t}),

with a branch of the logarithm chosen locally along the deformed contour. Differentiation gives

ϕ^{'} (z, t) = \frac{2 z}{z ^{2} - 1} - \frac{1}{z - t} = \frac{z ^{2} - 2 t z + 1}{( z ^{2} - 1 ) ( z - t )} .

For

t = cos θ

, its saddle points solve

z^{2} - 2 (cos θ) z + 1 = 0,

and hence

z_{\pm} = cos θ \pm i sin θ = e^{\pm i θ} .

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 336 / 1 / 1 / Solution Created 2026-09-24 Updated 2026-09-25

Write the Schläfli contour integral for Legendre polynomials as

P_{n} (μ) = \frac{1}{2 πi} \oint \frac{e ^{n ϕ (z)}}{2 ^{n} ( z - μ )} d z, ϕ (z) = lo g \frac{z ^{2} - 1}{z - μ} .

The saddle points satisfy

ϕ^{'} (z) = \frac{2 z}{z ^{2} - 1} - \frac{1}{z - μ} = 0, z^{2} - 2 μ z + 1 = 0.

For

μ = cos θ

, they are

z_{\pm} = e^{\pm i θ}

. Deforming the contour through both conjugate saddles and using the local Gaussian contributions from the method of steepest descent gives conjugate exponentials. Their sum is the Debye asymptotic for Legendre polynomials

P_{n} (cos θ) \sim \frac{2}{πn sin θ} cos [(n + \frac{1}{2}) θ - \frac{π}{4}] .

Both saddles are required for the real oscillatory answer.

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