这是一份kcl伦敦大学学院 6CCM326A作业代写的成功案例
$$
The second factor in is
$$
\begin{aligned}
\int_{0}^{\infty} d \ell \frac{\ell^{d-1}}{\left(\ell^{2}+\Delta\right)^{2}} &=\frac{1}{2} \int_{0}^{\infty} d\left(\ell^{2}\right) \frac{\left(\ell^{2}\right)^{\frac{d}{2}-1}}{\left(\ell^{2}+\Delta\right)^{2}} \
&=\frac{1}{2}\left(\frac{1}{\Delta}\right)^{2-\frac{d}{2}} \int_{0}^{1} d x x^{1-\frac{d}{2}}(1-x)^{\frac{d}{2}-1}
\end{aligned}
$$
where we have substituted $x=\Delta /\left(\ell^{2}+\Delta\right)$ in the second line. Using the definition of the beta function,
$$
\int^{1} d x x^{\alpha-1}(1-x)^{\beta-1}=B(\alpha, \beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}
$$
7CCMMS43 COURSE NOTES ：
Then we can write the currents associated with these symmetries as
$$
\begin{gathered}
j_{L}^{\mu}=\bar{Q}{L} \gamma^{\mu} Q{L}, \quad j_{R}^{\mu}=\bar{Q}{R} \gamma^{\mu} Q{R}, \
j_{L}^{\mu a}=\bar{Q}{L} \gamma^{\mu} \tau^{a} Q{L}, \quad j_{R}^{\mu a}=\bar{Q}{R} \gamma^{\mu} \tau^{a} Q{R},
\end{gathered}
$$
where $\tau^{a}=\sigma^{a} / 2$ represent the generators of $S U(2)$. The sums of left- and right-handed currents give the baryon number and isospin currents
$$
j^{\mu}=\bar{Q} \gamma^{\mu} Q, \quad j^{\mu a}=\bar{Q} \gamma^{\mu} \tau^{\alpha} Q
$$