# Dimensional regularizatoin of an integral

• MHB
• topsquark
In summary, the text suggests relabeling the coordinate of the integral as a complex number in order to regularize it.
topsquark
Science Advisor
Gold Member
MHB
This question hopefully isn't going to go too deep into the concept, just a couple of questions to get me going.

I am working on using dimensinal regularization of a loop integral in QED. I don't think the specific application to QED is important, but I will say that the original integral is in 4D Minkowsi space-time.

The example in the text starts with a Wick rotation of an integral in Minkowski space, changing the time coordinate by $$\displaystyle p^0 \to i p^4$$. The problem is now to integrate over 4D Euclidean space. The metric is therefore a Kronecker delta rather than the usual space-time metric but for some reason the text still uses the Minkowski metric, $$\displaystyle \eta ^{ \mu \nu }$$. In that spirit I'm just going to copy the text's notation.

So. To regularize the integral the text says
1) Relable the number of dimensions in the integral to d, which we will eventually take in the limit to 4. (I don't believe we need this comment.)

2) Drop all the terms in the integrand that are odd powers in p.

3) Replace the terms that are even powers of p with
$$\displaystyle p^{ \mu } p^{ \nu } \to p^2 \eta ^{ \mu \nu } / d$$

and
$$\displaystyle p^{ \mu } p^{ \nu } p^{ \rho } p^{ \sigma } \to (p^2)^2 \left ( \eta ^{ \mu \nu } \eta ^{ \rho \sigma } + \eta ^{ \mu \rho } \eta ^{ \nu \sigma } + \eta ^{ \mu \sigma } \eta ^{ \nu \rho } \right )/(d(d + 2))$$

1) I'm okay with this, though d is allowed to be complex. I really can't grok a complex "number" of dimensions but hey, I'm trying to evaluate a divergent integral. Welcome to the crazy world of QFT. (Party)

2) Why drop the odd powers?

3) Okay, the first one I can only work with but it doesn't seem too extreme. The second, I guess I could start out with $$\displaystyle p^{ \mu } p^{ \nu } p^{ \rho } p^{ \sigma } = \left ( p^{ \mu } p^{ \nu } \right ) \left ( p^{ \rho } p^{ \sigma } \right ) + \left ( p^{ \mu } p^{ \rho } \right ) \left ( p^{ \nu } p^{ \sigma } \right ) + \left ( p^{ \mu } p^{ \sigma } \right ) \left ( p^{ \nu } p^{ \rho } \right )$$
but the coefficient wouldn't work out. (These two substitutions bear a resemblance to the trace theorems of Dirac gamma matrices. (See pages 6 - 8.) Is there, perhaps, a reason for this?)

Given the starting points I can pretty much follow what is going on, though there is a Gamma function identity that I don't know how to derive. But other than that it's not really all that bad. I just can't figure out how to start it.

-Dan

topsquark said:
This question hopefully isn't going to go too deep into the concept, just a couple of questions to get me going.

I am working on using dimensinal regularization of a loop integral in QED. I don't think the specific application to QED is important, but I will say that the original integral is in 4D Minkowsi space-time.

The example in the text starts with a Wick rotation of an integral in Minkowski space, changing the time coordinate by $$\displaystyle p^0 \to i p^4$$. The problem is now to integrate over 4D Euclidean space. The metric is therefore a Kronecker delta rather than the usual space-time metric but for some reason the text still uses the Minkowski metric, $$\displaystyle \eta ^{ \mu \nu }$$. In that spirit I'm just going to copy the text's notation.

Hey Dan!

It's a bit problematic to have a coordinate that is a complex number.
As I understand it, only manifolds that map to real numbers are generally used. And metrics are typically defined to act on real numbers and produce a real number.

If we allow complex coordinates, we have a complex manifold, which means we now have an 8-dimensional space.
And our metric can produce complex numbers, meaning it doesn't properly measure curve length any more. That is, it is not actually a metric any more. It may look a little bit like an Euclidean metric, but it's not.
I think the general consensus in the relativity community is that although an imaginary time coordinate can give nicely simplified formulas in some cases, that it is not worth the trouble it brings.

Since it is mentioned in this context, I guess the text just uses it in some cases to show 'nice' formulas, but that it does not use this imaginary $p^4$ coordinate where it 'counts'.

topsquark said:
So. To regularize the integral the text says
1) Relable the number of dimensions in the integral to d, which we will eventually take in the limit to 4. (I don't believe we need this comment.)

2) Drop all the terms in the integrand that are odd powers in p.

3) Replace the terms that are even powers of p with
$$\displaystyle p^{ \mu } p^{ \nu } \to p^2 \eta ^{ \mu \nu } / d$$

and
$$\displaystyle p^{ \mu } p^{ \nu } p^{ \rho } p^{ \sigma } \to (p^2)^2 \left ( \eta ^{ \mu \nu } \eta ^{ \rho \sigma } + \eta ^{ \mu \rho } \eta ^{ \nu \sigma } + \eta ^{ \mu \sigma } \eta ^{ \nu \rho } \right )/(d(d + 2))$$

1) I'm okay with this, though d is allowed to be complex. I really can't grok a complex "number" of dimensions but hey, I'm trying to evaluate a divergent integral. Welcome to the crazy world of QFT. (Party)

Taking an integer dimension to a limit does sound strange.
Sounds more as if they first generalize to an arbitrary number of dimensions, and later on just set it to 4.

topsquark said:
2) Why drop the odd powers?

Dropping odd powers sounds like an application of anti-symmetry in a symmetric context.
Note that if $f$ is a symmetric function, then $\int_{-\infty}^\infty u^3\cdot f(u)\, du = 0$.

topsquark said:
3) Okay, the first one I can only work with but it doesn't seem too extreme. The second, I guess I could start out with $$\displaystyle p^{ \mu } p^{ \nu } p^{ \rho } p^{ \sigma } = \left ( p^{ \mu } p^{ \nu } \right ) \left ( p^{ \rho } p^{ \sigma } \right ) + \left ( p^{ \mu } p^{ \rho } \right ) \left ( p^{ \nu } p^{ \sigma } \right ) + \left ( p^{ \mu } p^{ \sigma } \right ) \left ( p^{ \nu } p^{ \rho } \right )$$
but the coefficient wouldn't work out. (These two substitutions bear a resemblance to the trace theorems of Dirac gamma matrices. (See pages 6 - 8.) Is there, perhaps, a reason for this?)

Given the starting points I can pretty much follow what is going on, though there is a Gamma function identity that I don't know how to derive. But other than that it's not really all that bad. I just can't figure out how to start it.

-Dan

Sounds as if a concept of rotational symmetry is applied, and we're only interested in an amplitude with respect to a specific origin.
The contribution of a specific point $(p_x, p_y)$ to the squared amplitude is $p^2=p_x^2+p_y^2$. Suppose amplitudes are equally distributed in all directions. If we consider the contributions of $p_x$ and $p_y$ separately, then the average contribution of each is $\frac 12p^2$.

Do you have such rotational symmetry where we can expect amplitudes in all directions?
And where we are only interested in some kind of average amplitude?

Klaas van Aarsen said:
Hey Dan!

It's a bit problematic to have a coordinate that is a complex number.
As I understand it, only manifolds that map to real numbers are generally used. And metrics are typically defined to act on real numbers and produce a real number.

If we allow complex coordinates, we have a complex manifold, which means we now have an 8-dimensional space.
And our metric might produce complex numbers, meaning it doesn't properly measure curve length any more. That is, it is not actually a metric any more. It may look a little bit like an Euclidean metric, but it's not.
I think the general consensus in the relativity community is that although an imaginary time coordinate can give nicely simplified formulas in some cases, that it is not worth the trouble it brings.

Since it is mentioned in this context, I guess the text just uses it in some cases to show 'nice' formulas, but that it does not use this imaginary $p^4$ coordinate where it 'counts'.
Taking an integer dimension to a limit does sound strange.
Sounds more as if they first generalize to an arbitrary number of dimensions, and later on just set it to 4.
Dropping odd powers sounds like an application of anti-symmetry in a symmetric context.
Note that if $f$ is a symmetric function, then $\int_{-\infty}^\infty u^3\cdot f(u)\, du = 0$.
Sounds as if a concept of rotational symmetry is applied, and we're only interested in an amplitude with respect to a specific origin.
If amplitudes are equally distributed in all directions, than the contribution of a specific point $(p_x, p_y)$ to the squared amplitude is $p^2=p_x^2+p_y^2$.
If we consider the contributions of $p_x$ and $p_y$ separately, then the average contribution of each is $\frac 12p^2$.

Do you have such rotational symmetry where we can expect amplitudes in all directions?
And where we are only interested in some kind of average amplitude?
Thanks for the comments. I'll look into this today and I'll post back some results if I can.

I really don't know much about your questions. The text talks about complex dimensions, but doesn't really explain any further. The end result is that the "infinite part" of the integral can be extracted from the integrand in the form of a divergent gamma function (for d = 4) and the limit is never actually taken. It is just considered to be an infinite constant and combined with another quantity, which is presumed to be infinite, and we can then "define away" the gamma function. (This is the part of the renormalization that is not really Mathematically sound and what people complain about.)

The original problem is Lorentz invariant and once the Wick rotation takes place we are left with an integral over all Euclidean space. The volume element used is described in the text as $$\displaystyle d^dp = \Omega _d \cdot p^{d - 1} dp$$, where $$\displaystyle \Omega _d$$ is the angular part of the d-hypersphere. (Come to think of it this also presumes that d is a positive integer.) In terms of the Euclidean version of the integral we are talking about spherical symmetry. I hadn't thought of it that way before.

I am guessing at this point that the regularization is already being applied to the given problem and the three initial steps that I was given may not represent the steps in a more general problem like I had thought.

Thanks for the input!

-Dan

topsquark said:
Thanks for the comments. I'll look into this today and I'll post back some results if I can.

I really don't know much about your questions. The text talks about complex dimensions, but doesn't really explain any further.

Complex dimensions make no sense to me either.

topsquark said:
The end result is that the "infinite part" of the integral can be extracted from the integrand in the form of a divergent gamma function (for d = 4) and the limit is never actually taken. It is just considered to be an infinite constant and combined with another quantity, which is presumed to be infinite, and we can then "define away" the gamma function. (This is the part of the renormalization that is not really Mathematically sound and what people complain about.)

I'm wondering how the gamma function is involved now... (Wondering)

topsquark said:
The original problem is Lorentz invariant and once the Wick rotation takes place we are left with an integral over all Euclidean space.

Just looked up Wick rotation and found that it is apparently a 'trick' to introduce an imaginary coordinate so that some problems are easier to solve. And then back substitute its solution into the actual problem to get results.

topsquark said:
The volume element used is described in the text as $$\displaystyle d^dp = \Omega _d \cdot p^{d - 1} dp$$, where $$\displaystyle \Omega _d$$ is the angular part of the d-hypersphere. (Come to think of it this also presumes that d is a positive integer.) In terms of the Euclidean version of the integral we are talking about spherical symmetry. I hadn't thought of it that way before.

Yeah, that seems to match the $dV = d^dp = r^2\sin\theta\,d\theta\,d\phi\,dr$ that we have in spherical coordinates. In the spherical case we'd have $\Omega_3 = \sin\theta\,d\theta\,d\phi$.

topsquark said:
I am guessing at this point that the regularization is already being applied to the given problem and the three initial steps that I was given may not represent the steps in a more general problem like I had thought.

I'm not familiar with regularization procedures for integrals covering divergent integrals... yet! (Cool)

Klaas van Aarsen said:
I'm wondering how the gamma function is involved now... (Wondering)
There are two integrals that come up in the process:
$$\displaystyle \int _0^{ \infty } dk ~ k^{d -1} \dfrac{k^2}{(k^2 + v^2)^2} = \dfrac{1}{2} (v^2)^{d/2 - 1} \Gamma \left ( 1 + \dfrac{d}{2} \right ) \Gamma \left ( 1 - \dfrac{d}{2} \right ) = \dfrac{1}{2} (v^2)^{d/2 - 1} \left ( \dfrac{d}{2} \right ) \Gamma \left ( \dfrac{d}{2} \right ) \Gamma \left ( 1 - \dfrac{d}{2} \right )$$

and
$$\displaystyle \int_0^{ \infty} dk ~ k^{d - 1} \dfrac{1}{(k^2 + v^2)^2} = \dfrac{1}{2} (v^2)^{d/2 - 2} \Gamma \left ( \dfrac{d}{2} \right ) \Gamma \left ( 2 - \dfrac{d}{2} \right ) = \dfrac{1}{2} (v^2)^{d/2 - 2} \left ( 1 - \dfrac{d}{2} \right ) \Gamma \left ( \dfrac{d}{2} \right ) \Gamma \left ( 1 - \dfrac{d}{2} \right )$$

I haven't quite figured out how to do them yet.

-Dan

Well, I'm out of ideas for solving those integrals by real methods. Tomorrow I hit the contour integrals.

-Dan

Okay. I was over at Physics Stack Exchange and someone was kind enough to provide me with most of the steps. It isn't actually all that hard, but it does involve a trick.

First, $$\displaystyle \Gamma (n) = \int _0^{ \infty } dt ~ t^{n - 1} e^{-t}$$. (Apologies in advance. I'm in Physics mode and we tend to write the differential element first.)

Note that $$\displaystyle \dfrac{1}{A^z} = \dfrac{1}{\Gamma (z)} \int_0^{ \infty } d \lambda ~ \lambda ^{z - 1} e^{-A \lambda }$$. We'll be needing z = 2 for this.

Off to the races. Let's work out
$$\displaystyle \int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{[k^2 + v^2]^2}$$

First let's simplify a bit; Let's do some factoring and let $$\displaystyle y = \dfrac{k}{v}$$. Then we get
$$\displaystyle \int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{[k^2 + v^2]^2} = v^{d - 4} \int_0^{ \infty } dy ~ y^{d - 1} \dfrac{1}{[y^2 + 1]^2}$$

Let $$\displaystyle A = y^2 + 1$$. From the above formula we have, then
$$\displaystyle \dfrac{1}{[y^2 + 1]^2} = \dfrac{1}{\Gamma (2)} \int_0^{ \infty } d \lambda ~ \lambda ^{2 - 1} e^{-(y^2 + 1) \lambda}$$

So
$$\displaystyle \int dk \text{ ... } = v^{d - 4} \int_0^{ \infty } \int_0^{ \infty} d \lambda ~ dy ~ \lambda ~ y^{d - 1} e^{-(y^2 + 1) \lambda }$$

$$\displaystyle \int dk \text{ ... } = v^{d - 4} \int_0^{ \infty } \int_0^{ \infty } d \lambda ~ dy ~ \lambda ~ y^{d - 1} e^{-y^2 \lambda} e^{- \lambda }$$

Now let $$\displaystyle \alpha = y^2 \lambda$$.

$$\displaystyle \int dk \text{ ... } = v^{d - 4} \int_0^{ \infty } \int_0 ^{ \infty } d \lambda ~ d \alpha ~ \dfrac{1}{2} ~ \alpha ^{-1/2} ~ \lambda ^{-1/2} \lambda \left ( \dfrac{ \alpha }{ \lambda } \right ) ^{(d - 1)/2} e^{- \alpha } e^{- \lambda }$$

Collecting terms and decoupling the integrals gives
$$\displaystyle \int dk \text{ ... } = \dfrac{1}{2} v^{d - 4} \left ( \int _0^{ \infty} d \alpha ~ \alpha ^{-1/2 + (d - 1)/2 } e^{- \alpha } \right ) \left ( \int_0^{ \infty } d \lambda ~ \lambda ^{-1/2 + 1 - (d - 1)/2} e^{ -\lambda } \right )$$

$$\displaystyle \int dk \text{ ... } = v^{d- 4} \left ( \int_0^{ \infty } d \alpha ~ \alpha ^{(d/2) - 1} e^{- \alpha } \right ) \left ( \int_0^{ \infty } d \lambda ~ \lambda ^{1 - (d/2)} ~ e^{- \lambda } \right )$$

So finally:
$$\displaystyle \int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{[k^2 + v^2]^2} = \dfrac{1}{2} v^{d - 4} \Gamma \left ( \dfrac{d}{2} \right ) \Gamma \left ( 2 - \dfrac{d}{2} \right )$$

Thanks for the help.

-Dan

Just to round out the thread I (finally) have an explanation of 3) from my original post. (It looks worse than it is.)

$$\displaystyle p^{ \mu } p^{ \nu } \to p^2 \eta ^{ \mu \nu } /d$$. The main idea here is that $$\displaystyle p^2$$ has to have the same value no matter what the number of dimensions are. So let's apply $$\displaystyle \eta _{ \mu \nu }$$ to both 4D and dD.
4D: $$\displaystyle \eta _{ \mu \nu } p^{ \mu } p^{ \nu } = p_{ \nu } p^{ \nu } = p^2$$

dD: $$\displaystyle \eta _{ \mu \nu } ~ p^2 \eta ^{ \mu \nu } = p^2 \eta _{ \mu \nu } ~ \eta ^{ \mu \nu } = p^2 \cdot d$$
So to get the same form for both we need to divide the RHS of the substitution by d, which explains the division by d on the RHS.

$$\displaystyle p^{ \mu } p^{ \nu } p^{ \rho } p^{ \sigma } \to (p^2)^2 \left ( \eta^{ \mu \nu } ~ \eta ^{ \rho \sigma } + \eta^{ \mu \rho } ~ \eta ^{ \nu \sigma } + \eta^{ \mu \sigma } ~ \eta ^{ \nu \rho } \right ) / (d(d + 2))$$

Here we need to have $$\displaystyle (p^2 )^2$$ on both sides. So let's multiply both sides by $$\displaystyle \eta _{ \mu \nu } ~ \eta _{ \rho \sigma }$$:
4D: $$\displaystyle \eta _{ \mu \nu } ~ \eta _{ \rho \sigma } p^{ \mu } p^{ \nu } p^{ \rho } p^{ \sigma } = p^2 \cdot p^2 = (p^2)^2$$
using the formula for 4D from above.

dD: Let's do this term by term.
$$\displaystyle (p^2)^2 \eta _{ \mu \nu } ~ \eta _{ \rho \sigma } ~ \eta ^{ \mu \nu } ~ \eta ^{ \rho \sigma } = (p^2)^2 \eta _{ \mu \nu } ~ \eta ^{ \mu \nu } ~ \eta _{ \rho \sigma } ~ \eta ^{ \rho \sigma } = (p^2)^2 \cdot d \cdot d = (p^2)^2 d^2$$$$\displaystyle (p^2)^2 \eta _{ \mu \nu } ~ \eta _{ \rho \sigma } ~ \eta ^{ \mu \rho } ~ \eta ^{ \nu \sigma } = (p^2)^2 \eta _{ \nu }^{ \rho} ~ \eta _{ \rho }^{ \nu } = (p^2)^2 \delta _{ \nu }^{ \rho } ~ \delta _{ \rho }^{ \nu } = (p^2)^2 \cdot d$$
where the $$\displaystyle \delta _{ \nu }^{ \rho }$$ is the Kronecker delta. (In fact you can replace all the $$\displaystyle \eta$$s with deltas for this as we are dealing with Euclidean dimensions.)

Similarly:
$$\displaystyle (p^2)^2 \eta _{ \mu \nu } ~ \eta _{ \mu \rho } ~ \eta ^{ \mu \sigma } ~ \eta ^{ \nu \rho } = (p^2)^2 \cdot d$$

Putting this all together:
$$\displaystyle \eta _{ \mu \nu } ~ \eta _{ \rho \sigma } (p^2)^2 \left ( \eta^{ \mu \nu } ~ \eta ^{ \rho \sigma } + \eta^{ \mu \rho } ~ \eta ^{ \nu \sigma } + \eta^{ \mu \sigma } ~ \eta ^{ \nu \rho } \right ) = (p^2)^2 ( d^2 + d + d) = (p^2)^2 d(d + 2)$$
so we need to divide by d(d + 2) in this case.

-Dan

## 1. What is dimensional regularization of an integral?

Dimensional regularization is a mathematical technique used to regulate divergent integrals in quantum field theory. It involves extending the number of dimensions in the integral from the usual three spatial dimensions to a higher number, typically four plus epsilon, where epsilon is a small number. This allows for the calculation of integrals that would otherwise be infinite.

## 2. Why is dimensional regularization used?

Dimensional regularization is used because it provides a more elegant and consistent way of dealing with divergent integrals in quantum field theory. It also allows for the use of renormalization, which is a method of removing infinities from calculations in order to obtain meaningful results.

## 3. How does dimensional regularization work?

Dimensional regularization works by replacing the usual three-dimensional space with a higher-dimensional space, which can be analytically continued to a complex number. This allows for the integration to be performed in a way that avoids the divergences that occur in the three-dimensional case.

## 4. What are the advantages of dimensional regularization?

One of the main advantages of dimensional regularization is that it leads to more consistent and well-behaved results compared to other regularization methods. It also allows for the use of renormalization, which is a powerful tool in quantum field theory. Additionally, dimensional regularization is a more elegant and mathematically rigorous approach to dealing with divergent integrals.

## 5. Are there any limitations to dimensional regularization?

While dimensional regularization is a powerful tool in quantum field theory, it does have its limitations. It can only be used in theories with a small number of dimensions, typically four or less. Additionally, it may not be applicable in certain situations, such as when dealing with non-renormalizable theories. In these cases, other regularization methods may need to be used.

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