In Chapter 12.2 of Weinberg's book, "The Quantum Theory of Fields", he gives a quick example of how differentiation of a divergent integral can be used to show that the divergence originates with constant terms. On page 506 of the text, he gives the result of the D=1 calculation, where D is the 'superficial degree of divergence' for the integration. In the following paragraph, he states that all of the constants of integration are divergent. However, when I evaluate the result at two different values of 'q' (the independent variable), I can solve for one of the constants in terms of purely finite terms. I have copied the TeX code for my analysis below.

Cancellation of Divergences (taken from Weinberg, Section 12.2)

Suppose we have a Feynman diagram with positive superficial degree of divergence, $D\geq 0$. Then, the part of the amplitude where all internal momenta go to infinity in the same way will diverge as: \begin{equation} \int^{\infty}k^{D-1}dk. \end{equation}

If we differentiate this $D+1$ times, we will then have a different expression with $D=-1$, which is convergent. As an example, we will use $D=0$, which is logarithmically divergent.

\begin{equation}\label{Divergent Example Def} \mathscr{I}(q) \equiv \int\limits_{0}^{\infty}\frac{dk}{k+q}. \end{equation}

Differentiating once, we have

\begin{equation} \mathscr{I}'(q) \equiv -\int\limits_{0}^{\infty}\frac{dk}{(k+q)^2} = -\frac{1}{q}, \end{equation}

which we can integrate to get

\begin{equation}\label{Divergent Example Integrated} \mathscr{I}(q) = -ln(q) + C. \end{equation}

Now, $q$ is finite and $C$ is a constant of integration. However, the two equations for $\mathscr{I}$ must be equal. Therefore, $C$ must diverge. The key assumption in this type of analysis is that the value of $C$ in this expression is the same for any value of $q$. This allows one to subract the $\mathscr{I}$ values from one another at different values of $q$ and arrive at a finite expression.

The same method can be used for an expression with $D=1$. The analysis is as follows.

\begin{align}\label{Divergent Example 2} \mathscr{I}(q) &\equiv \int\limits_{0}^{\infty}\frac{k}{k+q}dk\\ \mathscr{I}'(q) &= -\int\limits_{0}^{\infty}\frac{k}{(k+q)^2}dk\\ \mathscr{I}''(q) &= 2\int\limits_{0}^{\infty}\frac{k}{(k+q)^3}dk=\frac{1}{q}. \end{align}

This expression can be integrated twice to give (with intermediate integration shown explicitly):

\begin{align}\label{Divergent Example 2 Cont} \mathscr{I}'(q) &= ln(q) + C_1\\ \mathscr{I}(q) &= q\,ln(q)-q+C_1q+C_2\\ &\equiv q\,ln(q)+aq+b\label{D=1 Result Weinberg}. \end{align}

$a$ and $b$ are divergent constants in this expression. Again, these constants are the same for all values of $q$. Continuing in this manner, we can see that this method yields the following expression for $D>0$.

\begin{align}\label{Divergent Example General} \mathscr{I}(q) \sim \frac{q^D \, ln(q)}{D} + P(\mathscr{O}(D)), \end{align} where $P(\mathscr{O}(D))$ is a polynomial of order $D$. Each of the constants in the polynomial is divergent. There are $D+1$ of these divergent constants. Since these constants are divergent, we cannot simply measure $\mathscr{I}$ at $D+1$ values of $q$ to eliminate the unknowns.

However, if we know we can measure $\mathscr{I}$ at two values of $q$. Thus, if we subtract these two measured values, we should get the following.

\begin{align} \mathscr{I}_2 - \mathscr{I}_1 &\equiv \mathscr{I}(q_2) - \mathscr{I}(q_1)\\ &=q_2 ln(q_2) - q_1 ln(q_1)+a(q_2-q_1)\\ \Rightarrow a &= \frac{\left(\mathscr{I}_2 - \mathscr{I}_1\right) - \left[q_2 ln(q_2) - q_1 ln(q_1)\right]}{q_2-q_1}\label{Linear Constant}. \end{align}

The equation $\mathscr{I} \equiv q\,ln(q)+aq+b$ comes directly from Weinberg 12.2 (pg. 506). The result for $a$ seems to be a valid result if the expression for $\mathscr{I}(q)$ is valid. However, according to Weinberg, $a$ is divergent and my result seems to be finite for two distinct values of $q_1$ and $q_2$.

My question is: Why does it seem that I have a valid, finite result for $a$ when Weinberg says this term is divergent? Also, from my understanding, the divergent constants are at the core of the theory for perturbative renormalization.


1 Answer 1


The quick answer is $a$ diverges because $\mathcal{I}_2-\mathcal{I}_1$ diverges.

You have to be careful with the order of limits. For you to treat $\mathcal{I}_{1,2}$ like they are ordinary numbers you have to introduce some regulator, say just an ordinary cutoff at $k=\Lambda$. Then what you will get is $$\mathcal{I}(q)=q\log q+a(\Lambda) q+b(\Lambda) + \mathcal{O}(\frac{q}{\Lambda})$$ where $a,b$ are finite but $\Lambda$ dependent and diverge as $\Lambda\rightarrow \infty$. You can see this yourself by just simply evaluating your first integral with a finite upper limit, there is no need to take the $q$ derivatives to see this. You will also find some extra terms that go to zero as $\Lambda\rightarrow \infty$, but they aren't important here.

Now your argument is valid, and you do solve for a finite $a$ in terms of finite $\mathcal{I}_2-\mathcal{I}_1$, but both diverge as you send the cutoff to infinity.

  • $\begingroup$ @octonian, thanks for the explanation. My error seems to be that I thought the \mathscr{I} were measurable. I think I was confusing this explanation with something from a different part of the book. $\endgroup$
    – OldMan
    Nov 26, 2017 at 2:58
  • $\begingroup$ @OldMan, They are measurable in a sense. The modern view of renormalization is that the cutoff is something real, but we might not know exactly what the value of $\Lambda$ is, or how physical regularization happens. But the beauty of renormalizable theories is that for physics at $q<<\Lambda$ it doesn't matter how you regularize, so we can do something completely ad hoc like simply chopping off the integral at $\Lambda$. $\endgroup$
    – octonion
    Nov 26, 2017 at 3:19

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