Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 312 / 2 / iii

q = k_{1}

k - q = k_{2}

q = ∣ q ∣

, and

μ = k \cdot q

. Substitution into

F_{2} = 5 α_{s} /7 + 2 β /7

gives

F_{2} (q, k - q) = \frac{k ^{2} [ 7 k μ + q ( 3 - 10 μ ^{2} )]}{14 q ( k ^{2} - 2 k q μ + q ^{2} )} .

The azimuthal integral in

d^{3} q = q^{2} d q d μ d ϕ

and the factor two in

P_{22}

then give

P_{22} (k) = \int_{0}^{\infty} \frac{d q}{4 π ^{2}} \int_{- 1}^{1} d μ \frac{k ^{4} [ 7 k μ + q ( 3 - 10 μ ^{2} ) ] ^{2}}{98 ( k ^{2} - 2 k q μ + q ^{2} ) ^{2}} P (q) P (k^{2} - 2 k q μ + q^{2}) .

P (k) = A k^{n}

, put

r = q / k

and

s (r, μ) = 1 - 2 r μ + r^{2} .

All dimensional factors separate:

P_{22} (k) = A^{2} k^{2 n + 3} \int_{0}^{\infty} d r \int_{- 1}^{1} d μ K_{22} (r, μ; n),

where

K_{22} (r, μ; n) = \frac{r ^{n} [ 7 μ + r ( 3 - 10 μ ^{2} ) ] ^{2}}{392 π ^{2} [ 1 - 2 r μ + r ^{2} ] ^{, 2 - n /2}} .

r \to 0

K_{22} \sim r^{n} μ^{2} / (8 π^{2})

, so the soft-

q

integral converges exactly when

n > - 1

. For

r \to \infty

K_{22} \sim r^{2 n - 2}

times an angular function, requiring

n < 1/2

. At fixed

μ < 1

r \to 1

is regular. The corner

(r, μ) = (1, 1)

makes

∣ k - q ∣

soft; after resolving that corner in the local soft momentum, its radial behavior is again

p^{n} d p

, requiring

n > - 1

. Therefore

- 1 < n < \frac{1}{2}

is the infrared- and ultraviolet-convergence window for

P_{22}

Solved by gpt-5.6-sol high.

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