这是一份liverpool利物浦大学MATH421的成功案例

Now the coupling can be absorbed into the definition of this scale: Adding to the classical term $V_{0}=-\Lambda / g^{2}$, the total effective potential for $A$ up to one loop is
$$
V=-\Lambda\left[\frac{1}{g^{2}}+\ln \left(\frac{\Lambda}{\frac{1}{2} \mu^{2}}\right)-1\right]=-\Lambda\left[\ln \left(\frac{\Lambda}{\frac{1}{2} M^{2}}\right)-1\right]
$$
where $M$ is the “renormalization group invariant mass scale”:
$$
M^{2}=\mu^{2} e^{-1 / g^{2}}
$$

MATH421 COURSE NOTES ：

$$
u=(p \cdot x)^{2}-p^{2} x^{2}
$$
so we can evaluate the integral by either of the substitutions
$$
x^{2} \rightarrow 0, \quad p \cdot x \rightarrow \sqrt{u} \quad \text { or } \quad p \cdot x \rightarrow 0, \quad x^{2} \rightarrow-u / p^{2}
$$
We’ll consider now the latter choice. (The former gives the same result: See the exercise below.) Again, since we need to Taylor expand in $x$ anyway to find the result for a particular numerator, we expand and perform the $\lambda$ integration:
$$
\mathcal{A}{2}=\sum{n=0}^{\infty} \frac{1}{n !}\left(\frac{u}{2 p^{2}}\right)^{n}\left(\frac{1}{8} p^{2}\right)^{n+D / 2-2} \Gamma\left(2-\frac{D}{2}-n\right) \int_{0}^{1} d \beta\left(1-\beta^{2}\right)^{n+D / 2-2}
$$
Performing the change of variables $\beta^{2}=\gamma$ to convert the remaining integral to a Beta function, and using the identities
$$
\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}, \quad \Gamma(z) \Gamma(1-z)=\pi \csc (\pi z)
$$