# The way momentum space integrals tend to infinity

At the beginning of chapter 15 of Schwartz, he states that

$$\int d^4k \frac{k^2}{k^4}=\int \frac{d^4k}{k^2}\sim \int k\ dk.$$

I don't see how he got this at all. Isn't this just the integral

\begin{align*} \iiiint dwdxdydz \frac{1}{z^2 -(x^2 + y^2 +w^2)} &= \int dz \iiint drd\theta d\phi \ r^2\sin\theta \frac{1}{z^2-r^2} \\ &=4\pi\int\int_0^\infty dzdr \frac{r^2}{z^2 - r^2}\tag{1}\\ &=4\pi\int dz\ \left( z\tanh^{-1}\left(\frac{r}{z} \right) - r \right)_0^\infty\\ &\sim \left(-i\frac{\pi}{2}\right)\int dz\ z - \infty \end{align*}

No surprise, the integral is divergent, but the "first" infinity came from the $$r$$-integral, so so I'm kind of puzzled why he exactly pointed out the term $$\int dz\ z$$. Why doesn't he just go to polar coordinates and then point out that clearly what I've labeled as equation (1) is divergent? Is the infinity coming from the $$r$$-integral somehow irrelevant?

I feel like maybe I'm missing something in his logic or how he's going about solving the integrals, and I'm asking because he uses this "$$\sim$$" notation throughout his entire renormalization chapter, and I'm not exactly sure what he means.

You're trying to generalize 3D polar coordinates to 4D polar coordinates by introducing a "z-axis" in the fourth dimension. In some sense, you could consider it a generalization of cylindrical coordinates to 4D, in which every constant-z slice has a spherical (rather than circular) set of coordinates attached. This is all acceptable and works fine, but it's not how most people approach these integrals.

Most texts generalize the following line of reasoning:

In 3D, let $r$ be the magnitude of the position vector $(x,y,z)$. If we want to do an integral over all space that ONLY depends on $r$ (and not $x,y,z$ separately), then we can use polar coordinates to write

$$\int d^3\vec{r}\ f(r)=\int d\Omega dr\ r^2 f(r)=4\pi\int dr\ r^2 f(r)$$

where here $d\Omega$ is the angular part of the integral.

In 4D, we let $k$ be the magnitude of the position vector $(x,y,z,w)$. If we want to do an integral over all space that ONLY depends on r (and not $x,y,z,w$), then we can use polar coordinates to write

$$\int d^4\vec{k}\ f(k)=\int d\Omega dk\ k^3 f(k)=(\text{angular stuff})\int dr\ k^3 f(k)$$

The (angular stuff) you can look up online.

• I think the issue is that the integral over “angular stuff” is over a hyperbola, not a sphere, and is hence infinite. – knzhou Jul 9 '18 at 9:55
• @knzhou You might be right, I didn't fully think through the effect of the metric signature. I don't have a copy of Schwartz, but I assumed he was doing this integral after a Wick rotation, so the angular stuff is finite. – Jahan Claes Jul 9 '18 at 15:21
• Yeah. I think that’s the real issue OP was confused about. – knzhou Jul 9 '18 at 15:21

There are typically two approaches:

1. Integrate on z/$$x_0$$/t first and since there are poles in the propagator, you should device a way of going around the poles on the complex z plane, a la Feynman's $$i\epsilon$$ trick.
2. Do a Wick rotation of z to transition to the Euclidean metric. And then you can integrate via 4D polar coordinates without encountering any poles.

Either way you will arrive at the same quadratic divergence results.

Or you can rightfully and perversely insist on third approach:

1. Integrate on 3D polar coordinates first. But you should still have to properly handel the poles in the propagator, instead of just ignoring the poles as in your OP. If you do the poles justice, you will arrive at the same results as the other approaches.