If $\alpha_{s}\left(Q^{2}\right)=c / \log Q^{2}$, show that (10.37) leads to
$$
\frac{\int x^{n-1} q\left(x, Q^{2}\right) d x}{\int x^{n-1} q\left(x, Q_{0}^{2}\right) d x}=\left(\frac{\log Q^{2}}{\log Q_{0}^{2}}\right)^{A_{n}}
$$
where
$$
\begin{aligned}
A_{n} &=\frac{c}{2 \pi} \int_{0}^{1} x^{n-1} P_{q q}(x) d x \\
&=\frac{c}{2 \pi} \frac{4}{3}\left(-\frac{1}{2}+\frac{1}{n(n+1)}-2 \sum_{j=2}^{n} \frac{1}{j}\right)
\end{aligned}
$$
That is, in QCD, the moments $(n \geq 1)$ of the quark structure functions decrease as calculable powers of $\log Q^{2} \cdot c$ is given by (7.65).
The observant reader may have noticed what appears to be a contradiction in our interpretation of the $P$ functions. For example, we regard the $P_{q q}$ term as the correction factor to the quark density that arises from allowing for gluon emission. However, there are two diagrams: one with the gluon emitted from the initial quark line and the other with the gluon radiated from the final quark line; see Fig. 10.2. Our picture is only valid if the first diagram dominates. Then, the emitted gluon can be considered as part of the proton structure. It is a "partonlike" diagram. It turns out that both diagrams are required to ensure gauge invariance of the amplitude, but that the second only plays the role of canceling the contributions from the unphysical polarization states of the gluon. Adopting a physical gauge, in which we sum only over transverse gluons, only the first diagram remains.