# Taming Infinities

1. Feb 14, 2008

### robousy

I'm trying to work out the steps in the following:

$$-\frac{1}{2}\int\frac{d^{2n}k}{(2\pi)^{2n}}\frac{1} {\Gamma(s)}\sum_{m=-\infty}^{m=\infty}\int_0^\infty t^{s-1}e^{-(k^2+a^2m^2)t}dt=-\frac{\pi^n}{(2\pi)^{2n}L^{2n}}\frac{\Gamma(s-n)}{\Gamma(s)}L^{2s}\zeta(2s-2n)$$

I've tried a number of things including using the Jacobi theta function to change the limits on the summation, integral representation of the modified bessel function, and various expressions for the gamma and zeta functions, but to no avail. I've spent about 3 days on this so far. If anyone experienced in regularization can help me get from the left hand side to the right hand side I will be forever grateful!

Richard

Last edited: Feb 14, 2008
2. Feb 19, 2008

### blechman

This integral isn't quite as bad as it looks. You certainly don't need Jacobi theta functions! In what follows I am assuming that $L\equiv a^{-1}$ otherwise the expression does not make sense dimensionally.

First, notice that many things cancel out right from the start! For example, all of the pi's go away (recall that in D=2n, the measure is $d^{2n}k=(\pi^n/\Gamma(n))x^{n-1}dx$ where $x\equiv k^2$). In addition, since the quantity beings summed is even, we can use the factor of 1/2 to sum from 1 to infinity, with the zero term being left over. Putting this all together, and letting $y=k^2/a^2$, leaves me with:

$$\frac{\Gamma(s)}{\Gamma(n)}\int_0^\infty dy\left[\frac{1}{2}y^{n-s-1}+\sum_{m=1}^\infty (y+m^2)^{-s}y^{n-1}\right]=\Gamma(s-n)\zeta(2s-2n)$$

The first term vanishes in DimReg, so that we need only consider the term with the sum. We can factor out the $m^2$ from this and let $z=y/m^2$ to obtain:

$$\frac{\Gamma(s)}{\Gamma(n)}\int_0^\infty dz\left(\sum_{m=1}^\infty \frac{1}{m^{2s-2n}}\right)(1+z)^{-s}z^{n-1}$$

The quantity in parentheses is just $\zeta(2s-2n)$ and the final integral can be experssed in terms of Gamma functions and the identity is proven. QED.

3. Mar 2, 2008

### robousy

Blechman, thank you for the reply and apologies for the delay, I didn't think anyone had written back on this.

I really like this proof. I had finally managed to figure out a slightly different route but your method has demonstrated that there is more than one way to skin a cat.

One thing that did confuse me was the term:

$$\int_0^\infty dy \frac{1}{2}y^{n-s-1}$$

which you mentioned vanished in dimensional regularization. I'm having trouble seeing that as it looks pretty divergent.

Rich

4. Mar 2, 2008

### kdv

In dimensional regularization all power law divergent integral give zero.

5. Mar 2, 2008

### blechman

quite so. All scale-less integrals vanish in dim-reg. The "stupid" way to see this is that all the answers in dim reg go as log of some scale. If there is no scale, then there is nothing to take the log of, and hence the result must vanish. If you are still troubled: the idea is that the UV-divergence and the IR-divergence cancel. You can see this in several ways:

(1) Break the integral up: [0,infty] = [0,a]+[a,infty]. The first integral diverges at the lower endpoint (so for $n=2-\epsilon$ you must have $\epsilon<0$ to regulate), but the second integral diverges at the upper endpoint, so it comes with $\epsilon>0$. They both diverge like $1/\epsilon$ so the two divergences cancel!

(2) Put in an explicit IR regulator $y\rightarrow y+a$. After doing the integral but before you expand in $\epsilon$, notice that the answer is proportional to $a^p$ where p is a function of $\epsilon$. Now take $a\rightarrow 0$ BEFORE you do the $\epsilon$ expansion. You see that the answer vanishes. Of course, you will need to make sure you chose $\epsilon$ so that p>0 for this to work; see method (1).

In conclusion, the key point is that in dim-reg you have turned the integral into a meromorphic function of dimension, and can therefore move around in "$\epsilon$ space". This also leads to the wonderful result that dim reg is only sensitive to LOGARITHMIC divergences. This is good news, since in QFT, these are the only divergences that matter! That is, they're the divergences that make up the "renormalization group".

Hope this helps!

6. Mar 3, 2008

### robousy

Hey Blech,

Well, I'm probably going to reveal some profound level of mathematical ignorance here, but I'm working though the 1st set of instructions in your post above.

1. Break the integral up

$$\int_0^\infty dy \frac{1}{2}y^{n-s-1}=\int_0^a dy \frac{1}{2}y^{n-s-1}+\int_a^\infty dy \frac{1}{2}y^{n-s-1}$$

Perform integration:

$$\left[\frac{y^{n-s}}{n-s}\right]_0^a+\left[\frac{y^{n-s}}{n-s}\right]_a^\infty$$

Inserting values:

$$0+\frac{a^{n-s}}{n-s}-\frac{a^{n-s}}{n-s}+\frac{\infty^n-s}{n-s}=\infty$$

This isn't too illuminating. I'm sure I'm looking at this in a very naive way, but what steps in your unstructions am I misunderstanding?

Rich

7. Mar 3, 2008

### blechman

Consider the integral:

$$I=\int_0^\infty \frac{dx}{x^{1+a}}$$

This integral diverges, obviously. If a>0, it diverges at the lower limit; if a<0 it diverges at the upper limit; if a=0 it diverges at both places.

Performing the integral gives:

$$I=\left(\frac{1}{a}\lim_{x\rightarrow 0}\frac{1}{x^a} - \frac{1}{a}\lim_{x\rightarrow\infty}\frac{1}{x^a}\right)$$

The first term vanishes if we let a<0, and the second term vanishes if we let a>0. Thus I=0!

Again, the whole point is that we can do the upper and lower limits in different dimensions! Remember: a is a function of $\epsilon$. So in what you did below, you have to chose your n's carefully!

8. Mar 3, 2008

### robousy

Thats really insightful bletchman! I've been working through your explanation and its all making a lot more sense. So this is really interesting, your saying that for dim reg to work you have to give freedom to the value of $\epsilon$ that necessarily changes from the lower integral bound to the upper integral bound.

Is that 'alllowed' ?? Obviously so, but it seems a litte fishy. I'm going to read up a litte more on the meromorphic functions that you mentioned as I don't have any experiences with them.

Are you saying that it only works when a function diverges logarithmically? Also, the integrals above are only logs if the term 'a' is zero right? eg, $$I=\int_0^\infty \frac{dx}{x^{1+a}}$$ becomes logarithmic if a=0, but this seems like a special case.

Apologies in advance if this all sounds tediously ignorant!

Thanks again for your insights on this, its really useful.

Rich

9. Mar 4, 2008

### blechman

of course it's not ALLOWED - Just like any other regulator!

seriously, it's fine as long as you're willing to accept the regulator. Dim-Reg, although very (let me say again, VERY!) useful in practice, can do a great deal to muddle up what's really going on. My advice (and this comes from some of the best QFT experts out there): if you REALLY want to see what's going on, use Pauli-Villars! If you just want a quick and simple answer, use DimReg.

meromorphic == analytic up to a bunch of isolated singularities (poles).

sorry, maybe I wasn't very clear. The point is that dim reg is completely insensitive to power-law divergences, such as the quadratic divergence of a scalar field mass. DimReg only sees the log divergence. This is just because, due to the above argument, integrals that have power-law divergences vanish in dim-reg. Only the log divergences remain.

And this is good news, since the log divergences have a "physical" interpretation in terms of the renormalization group, whereas the power-law divergences are unphysical. So once again, DimReg has proved its usefulness by only keeping the divergences we want!