# Cutting off divergent integrals

In Quantum Field Theory, one has to deal with loop integrals which are divergent. The way out is to regularise your integral. For example one can introduce a cutoff, like in the Pauli-Villlars regularisation scheme.

I was trying to learn this method, and I found some useful material in Zee's book: Quantum Field Theory in a Nutshell. But I'm having some math-related difficulty, and I hope someone could offer some help.

In short words, once we find a divergent integral (in QFT they appear in a common form), we can exploit the fact that the following integral, which sometimes is called a master formula, is convergent,

$$\int \frac{d^4 k}{(2 \pi)^4} \frac{1}{(k^2 - c^2 + i \epsilon)^3} = \frac{- i}{32 \pi^2 c^2}$$

Where, k is a 4-momentum, and c is usually a mass.

Then, if we face a divergent integral, like (we can see it by power counting),

$$\int \frac{d^4 k}{(2 \pi)^4} \frac{1}{(k^2 - c^2 + i \epsilon)^2}$$

We first introduce a cutoff in the Pauli-Villars way, which is to replace the integrand by:

$$\int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}]$$

Where, $\Lambda$ is the cutoff. And we are taking very large k and $\Lambda^2 >> c^2$.

In order to calculate this integral with this cutoff, I first differentiate the "divergent" integral wrt $c^2$ until I find a convergent integral. So,

$$\frac{d}{dc^2} \int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \int \frac{d^4 k}{(2 \pi)^4} \frac{2}{(k^2 - c^2 + i \epsilon)^3} = \frac{- i}{16 \pi^2 c^2}$$

Then, I integrate back,

$$\int d \int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \int dc^2 \frac{- i}{16 \pi^2 c^2}$$

Which gives (according to me :p and here where I think I'm missing something),

$$\int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \frac{- i}{16 \pi^2} log(c^2)$$

$$\int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \frac{i}{16 \pi^2} log(\frac{\Lambda^2}{c^2})$$

Similarly for other integrals, such as

$$\int \frac{d^4 k}{(2 \pi)^4} [\frac{k^2}{(k^2 - c^2 + i \epsilon)^2} - \frac{k^2}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \frac{-i}{16 \pi^2} [\Lambda^2 - 2c^2 log(\frac{\Lambda^2}{c^2}) + c^2 + ...]$$

Which I don't know how to reach with my calculation. Is it that when we integrate back, we have the freedom to add a suitable constant (like $+ log(\Lambda^2)$ for the first case, and $+\Lambda^2 + log(\Lambda^2)$ for the second)? Or something else?

I hope the question is clear and someone will help me out here.

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you should probably fix the error in your first integral... as it stands now it diverges quadratically.... – Kyle Nov 28 '11 at 1:18
Oh! Thanks for mentioning the typo. It's fixed now. – stupidity Nov 28 '11 at 1:23

I think the answer is that you can split up your integral into two identical integrals (up to a sign), you know the answer to both, the one with $c^2$ comes with a $-$ and the one with $\Lambda^2$ comes with a $+$ sign. Just make sure when you are integrating back after taking the derivative that you take the same bottom limit i.e.
$\int_{Const.}^{c^2} f'(s)ds$ and $\int_{Const.}^{\Lambda^2} f'(s) ds$ where $f(x)$ is the integral
$f(x) = \int \frac{d^4 k}{(2 \pi)^4} \; \frac{1}{(k^2-x^2 + i \epsilon)^2}$
Thanks for your answer. That's what I did first, and I got a correct result for the log divergent integral. But for the quadratically divergent one, I get something like $~ \Lambda^2 - \frac{\Lambda^2}{c^2}log{\Lambda^2}{c^2} + c^2$ But this expression has the wrong dimension for the middle term (it's dimensionless), whereas $\Lambda^2$ and $c^2$ are of $(mass)^2$ dimension. The correct answer for the middle term should be something like $~c^2log\frac{\Lambda^2}{c^2}$ – stupidity Nov 28 '11 at 14:06
There's a typo in the previous comment. the middle term in the first expression reads, $\frac{\Lambda^2}{c^2} log(\frac{\Lambda^2}{c^2})$ – stupidity Nov 28 '11 at 14:13