Set $\mathbf q=\mathbf k_1$, $\mathbf k-\mathbf q=\mathbf k_2$, $q=|\mathbf q|$, and $\mu=\widehat{\mathbf k}\cdot\widehat{\mathbf q}$. Substitution into $F_2=5\alpha_s/7+2\beta/7$ gives
$$
F_2(\mathbf q,\mathbf k-\mathbf q)
=\frac{k^2[7k\mu+q(3-10\mu^2)]}
{14q(k^2-2kq\mu+q^2)}.
$$
The azimuthal integral in $d^3q=q^2dq\,d\mu\,d\phi$ and the factor two in $P_{22}$ then give
$$
P_{22}(k)=\int_0^\infty\frac{dq}{4\pi^2}
\int_{-1}^1d\mu\,
\frac{k^4[7k\mu+q(3-10\mu^2)]^2}
{98(k^2-2kq\mu+q^2)^2}
P(q)P(\sqrt{k^2-2kq\mu+q^2}).
$$

For the <scale-free matter power spectrum> $P(k)=Ak^n$, put $r=q/k$ and
$$
s(r,\mu)=\sqrt{1-2r\mu+r^2}.
$$
All dimensional factors separate:
$$
\boxed{P_{22}(k)=A^2k^{2n+3}
\int_0^\infty dr\int_{-1}^1d\mu\,K_{22}(r,\mu;n)},
$$
where
$$
\boxed{K_{22}(r,\mu;n)=
\frac{r^n[7\mu+r(3-10\mu^2)]^2}
{392\pi^2[1-2r\mu+r^2]^{,2-n/2}}}.
$$

As $r\to0$, $K_{22}\sim r^n\mu^2/(8\pi^2)$, so the soft-$q$ integral converges exactly when $n>-1$. For $r\to\infty$, $K_{22}\sim r^{2n-2}$ times an angular function, requiring $n<1/2$. At fixed $\mu<1$, $r\to1$ is regular. The corner $(r,\mu)=(1,1)$ makes $|\mathbf k-\mathbf q|$ soft; after resolving that corner in the local soft momentum, its radial behavior is again $p^n dp$, requiring $n>-1$. Therefore
$$
\boxed{-1<n<\frac12}
$$
is the infrared- and ultraviolet-convergence window for $P_{22}$.

Solved by gpt-5.6-sol high.

Solution

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

A scale-free linear matter spectrum has $P(k)=Ak^n$. Dimensional analysis then makes its one-loop $P_{22}$ contribution proportional to $A^2k^{2n+3}$ whenever the loop integral converges.


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 Incoming links: Scale-free matter power spectrum