3
$\begingroup$

I am currently leaning the renormalisation of QED and I have met some tricky integral that seems unsolvable. The integrals are shown in Quantum Field Theory and the Standard Model by Schwartz, page 348, first edition.The integral (19.55) can be split into 2 parts: $$\int_0^1 \frac{{m_R}^2(1-z) \left(z^2-4 z+1\right)}{m_{\gamma} z+{m_R}^2 (1-z)^2} \, dz\tag{19.55} $$

And $$ \int_0^1 \ln \frac{{\mu}^2}{(1-z)^2{m_R}^2+z{m_{\gamma}}^2}\,dz.\tag{19.55} $$ The two integral altogether give results $$ \frac{1}{2}log\frac{{\mu}^2}{{m_R}^2}+3+log\frac{{m_\gamma}^2}{{m_R}^2} $$

Similar integrals can also be seen on Page 358, (20.16) $$ \int_0^1 \,dx\int_0^{1-x}\,dy [\ln\frac{(1-x-y){\Lambda}^2}{-xyQ^2+(1-x-y){m_{\gamma}}^2}]=\frac{3}{4}+\frac{1}{2}log\frac{{\Lambda}^2}{-Q^2} +\mathcal{O}(m_\gamma) \tag{20.16} $$ And on Page 359, (20.17) $$ \int_0^1 \,dx \int_0^{1-x}\,dy \frac{Q^2(1-x)(1-y)}{-xyQ^2+(1-x-y){m_{\gamma}}^2}=-\frac{1}{2}log^2\frac{{m_\gamma}^2}{-Q^2-i\epsilon}-2\frac{{m_\gamma}^2}{-Q^2-i\epsilon}-\frac{{\pi}^2}{3}-\frac{5}{2}+\mathcal{O}(m_\gamma)\tag{20.17} $$ The author gave us the final result without any computations. $\ $

But all these integrals seem unsolvable as they possess integral variable both on the numerator and the denominator. For the second an third integral I show here even involves logarithm which is supposed to make it more complicated. $ \ $

So is there any methodology that can help us compute these integral manually or is there any approximation that I miss in the process?

$\endgroup$
0

2 Answers 2

2
$\begingroup$

Your suspicion is justified. You are indeed missing a crucial approximation not mentioned in your post, turning a simple few-line calculation into an exercise in self-torture. The photon mass $m_\gamma$, present in eq. (19.55) of your book, is an infrared regulator to be considered only in the limit $m_\gamma \to 0$.

The simpler of the two integrals in eq. (19.55), $$I_1= \int\limits_0^1 \! dz \, (1-z) \ln \frac{\tilde{\mu}^2}{(1-z)^2m_R^2+zm_\gamma^2}, \tag{1} \label{1}$$ is even infrared finite. We can safely put $m_\gamma=0$, obtaining the result $$I_1=\frac{1}{2} \left( \ln \frac{\tilde{\mu}^2}{m_R^2}+1\right) \tag{2} \label{2}$$ after a short calculation.

The integral $$I_2=\int\limits_0^1 \! dz\, (1-z)\frac{m_R^2(1-4z+z^2)}{(1-z)^2m_R^2+zm_\gamma^2} \tag{3} \label{3}$$ is indeed infrared divergent and we have to work a bit more. For the following calculation, it is slightly more convenient to recast the integral into the form $$I_2=\int\limits_0^1 \! dx\, \frac{x(x^2+2x-2)}{x^2+(1-x)\lambda^2}, \quad \text{with} \quad \lambda= m_\gamma/m_R, \tag{4} \label{4}$$ where the transformation of variables $x=1-z$ was performed in \eqref{3}. Using $$\lim\limits_{\lambda \to 0} \int\limits_0^1 \! dx \left(\frac{x(x^2+2x-2)}{x^2+(1-x)\lambda^2}+ \frac{2x}{x^2+\lambda^2} \right) =\int\limits_0^1 \! dx \, (2+x) =\frac{5}{2}, \tag{5} \label{5} $$ and $$\int\limits_0^1 \! dx \, \frac{2x}{x^2+\lambda^2}=\int\limits_{\lambda^2}^{1+\lambda^2} \frac{du}{u}=\ln \frac{1+\lambda^2}{\lambda^2}=-\ln \lambda^2+\mathcal{O}(\lambda^2), \tag{6} \label{6}$$ we find $$I_2= \ln \lambda^2 +\frac{5}{2}= \ln \frac{m_\gamma^2}{m_R^2}+\frac{5}{2}. \tag{7} \label{7}$$ Adding \eqref{2} and \eqref{7}, we obtain $$I_1+I_2= \frac{1}{2} \ln \frac{\tilde{\mu}^2}{M_R^2} + \ln \frac{m_\gamma^2}{m_R^2}+3, \tag{8}$$ obviously in agreement with the result given in the second line of eq. (19.55) of the book by Schwartz.

$\endgroup$
1
$\begingroup$

Your complaint is not entirely justified, the first integrand is just a rational function, for which solution methods can easily be found: integration-rational-functions Although tedious, it can be done by hand.

Still the use of a symbolic algebra engine (like downloadable Wolfram Engine ) would of course be a big help. Often this works best if we first scale the integrals to make the expressions simpler. Doing this for the second expression, for instance, would lead to:

In[1]:= f=Log[a z+(1-z)^2]
                   2
Out[1]= Log[(1 - z)  + a z]

In[2]:= Iz=Integrate[f, z, Assumptions->{a>0}]

                                             -2 + a + 2 z              a                            2
Out[2]= -2 z + Sqrt[-((-4 + a) a)] ArcTan[-------------------] + (-1 + - + z) Log[1 + (-2 + a) z + z ]
                                          Sqrt[-((-4 + a) a)]          2

In[3]:= Limit[Iz,z->0]
                                         -2 + a
Out[3]= Sqrt[-((-4 + a) a)] ArcTan[-------------------]
                                   Sqrt[-((-4 + a) a)]


In[4]:= Limit[Iz,z->1]
                                                 a             a Log[a]
Out[4]= -2 + Sqrt[-((-4 + a) a)] ArcTan[-------------------] + --------
                                        Sqrt[-((-4 + a) a)]       2

You would then have to combine the limits of the primitive function and re-scale the result by putting back the factors that we left out, or you could try if the engine can solve the "dressed up" case directly, but for the most difficult cases that often fails, or takes a very long time, or gives an excessively long expression as a result. So in practice, a trial-and-error combination of automated computation and redirections by hand is often the best solution.

As a further demonstration, let's look at the third case. Without loss of generality it can be changed to an expression with just one constant $a$: $$ \int_0^1 \,dx\int_0^{1-x}\,dy\ \ln\big( \frac{1-x-y}{-a\,xy\ +\ 1-x-y} \big), \tag{20.16 simpler} $$ where the difference with the original is just the value of $\int_0^1 dx\int_0^{1-x}dy\, \ln(\Lambda^2/m_\gamma^2)$ and the new parameter is $a= Q^2/m_\gamma^2$. If we split, just for clarity, the computation in two steps, we then obtain without further problems:

In[1]:= f=Log[(1-x-y)/(1-x-y - a x y) ]

                1 - x - y
Out[1]= Log[-----------------]
            1 - x - y - a x y

In[2]:= Iy=Integrate[f, {y,0,1-x}, Assumptions->{a>0,x>0,x<1}]

        a (-1 + x) x (-I Pi + Log[a] + Log[x])
Out[2]= --------------------------------------
                       1 + a x

In[3]:= Integrate[Iy, {x,0,1}, Assumptions->{a>0}]
                                                          2
Out[3]= (3 a (4 + 3 a) + (6 I) a (2 + a) Pi - 2 (1 + a) Pi  - 6 a (2 + a) Log[a] +

                                                                               1            2
>      6 (1 + a) Log[1 + a] ((-2 I) Pi + Log[1 + a]) + 12 (1 + a) PolyLog[2, -----]) / (12 a )
                                                                             1 + a

If, on the other hand, we would just enter the expression straightforwardly, without offering the engine enough help, you might get: (after a long waiting time)

In[1]:= f=Log[Lambda^2(1-x-y)/(mg^2(1-x-y) - Q^2 x y) ]

                    2
              Lambda  (1 - x - y)
Out[1]= Log[------------------------]
              2                2
            mg  (1 - x - y) - Q  x y

In[2]:= Integrate[ Integrate[f, {y,0,1-x}],  {x,0,1}, Assumptions->{mg>0,Lambda>0,Q>0}]

Out[2]= Undefined

So a little bit of extra work by hand is usually needed!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.