# Explicit computation of singular part of two-loop sunrise diagram

For $$\phi^4$$, there is two-loop self-energy contribution from sunrise (sunset) diagram.

The integration is $$I(p)=\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D}\frac{1}{(p_1^2+m^2)(p_2+m^2)[(p-p_1-p_2)^2+m^2]}.$$

I try to use trick like Feynman Parameter, but still cannot get the explicit result. I saw this question Two-loop regularization and I found that the exact solution of above integration is very complex and need a lot of fancy technique. However in order to do the renormalization I need only singular part of $$I(p)$$ ($$\epsilon$$-expansion with $$D=4-\epsilon$$).

My question:

1. Is there some easy and direct method just to compute the singular part of $$I(p)$$ i.e. the coefficient of $$1/\epsilon^2$$ and $$1/\epsilon$$?

By the way, I saw this answer https://physics.stackexchange.com/a/79236/169288 But it only give an explicit computation of singular of $$I^\prime(p^2=m^2)$$ which is not my requirement.

You just need to provide the literature or textbooks which show the explicit computation details. I can rarely find textbooks covering explicit two-loop computation.

PS: The following is my method by ordinary Feynman parameter trick.

$$I(p)=\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D}\frac{1}{(p_1^2+m^2)(p_2+m^2)[(p-p_1-p_2)^2+m^2]}$$ $$I(p)=\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \int dx dy dz \delta(x+y+z-1) \frac{2}{\mathcal{D}^3}$$ with
$$\mathcal{D}=x p_1^2 +y p_2^2 +z(p-p_1-p_2)^2 +m^2 = \alpha k_1^2 + \beta k_2^2 + \gamma p^2 +m^2$$ with $$\alpha = x+z$$ $$\beta = \frac{xy + yz +zx}{x+z}$$ $$\gamma = \frac{xyz}{xy + yz +zx}$$ $$k_1= p_1 + \frac{z}{x+z}(p_2-p)$$ $$k_2 =p_2 - \frac{xz}{xy + yz +zx} p.$$

And the Jacobian $$\frac{\partial(k_1,k_2)}{\partial(p_1,p_2)}=1.$$

$$I(p)=\int_0^1 dx dy dz \delta(x+y+z-1) \int\frac{d^D k_1}{(2\pi)^D}\frac{d^Dk_2}{(2\pi)^D} \frac{2}{(\alpha k_1^2 + \beta k_2^2 + \gamma p^2 +m^2 )^3}$$ $$I(p)=\int dx dy dz \delta(x+y+z-1) \int\frac{d^D k_1}{(2\pi)^D}\frac{d^Dk_2}{(2\pi)^D} \int_0^{+\infty} dt t^2 e^{-t(\alpha k_1^2 + \beta k_2^2 + \gamma p^2 +m^2) }.$$ Gaussian integral of $$k_1$$ and $$k_2$$ $$I(p)= \int dx dy dz \delta(x+y+z-1)\int_0^{+\infty} dt \frac{t^{2-D}}{ (4\pi)^D (\alpha \beta)^{D/2}}e^{-t(\gamma p^2 +m^2)}$$ $$I(p)=\int dx dy dz \delta(x+y+z-1) \frac{\Gamma(3-D)}{(4\pi)^D(\alpha \beta)^{D/2}(\gamma p^2 +m^2)^{3-D}}.$$ Use $$D= 4-\epsilon$$ $$I(p)=\frac{1}{(4 \pi)^4}\int dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\Gamma(-1+\epsilon) \left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right)^\epsilon$$

$$\Gamma(-1+\epsilon)= -\frac{1}{\epsilon} +\gamma_E-1 +\mathcal{O}(\epsilon)$$ $$\left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right)^\epsilon = 1 + \epsilon \ln\left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right)+\mathcal{O}(\epsilon^2).$$

Up to $$0$$th order of $$\epsilon$$ $$I(p)=\frac{1}{(4 \pi)^4}\int dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\left(-\frac{1}{\epsilon}- \ln\left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right) +\gamma_E-1\right) .$$

There is two integral I need to compute $$I_1 = \int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}$$ $$I_2 = \int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right).$$ I can't find the explicit result of these two.

• @InertialObserver Not the dimensional regularization doesn't work. I can't find the explicit formula of singular part. You can see the website cited above. Commented Jan 1, 2019 at 5:59
• Right but if you just follow the usual procedure expanding each in $\frac{1}{\epsilon}$.. I certainly think it's doable just with regular methods.. the question is if I want to go through it Commented Jan 1, 2019 at 6:01
• For the innermost integral you can use Feynman parameters and just proceed as usual.. the question is what comes after that. I recommend at least doing that much so that you only have one integral to tackle.. I would be more inclined to help at that point Commented Jan 1, 2019 at 6:04
• But to answer your question. No, there is no "easy" way to extract the divergent part. You've got to go through it.. That's what makes higher order corrections so difficult Commented Jan 1, 2019 at 6:07
• I find this two-loop self-energy diagram calculated using integration by parts(IBP) in p115, Critical Properties of Phi4-Theories by Kleinert.H. This book also give some methods to higher loop diagram. A general discussion of IBP could be found in Analytic Tools for Feynman Integrals by A. Smirnov. Commented Apr 9, 2019 at 15:48

If you can be satsfied with the $$m=0$$ case, the integral is easy provided you work in configuaration space rather that momentum space. The $$x$$-space propagator in $$n$$ dimensions is $$g(x,x') = \frac{1}{ (n-2)S_{n-1}} \left(\frac {1}{|x-x'|}\right)^{n-2}$$ where $$S_{n-1} = 2\pi^{n/2}/\Gamma(n/2)$$ is the surface area of the $$n$$-ball. Your Feynman diagram in configuration space is $$I(p)= \int {d^nx} e^{ipx}[g(x,0)]^3$$ which you can evaluate (after re-naming the following integrals integration variable $$k$$ as $$x$$ and its $$x$$ to the external momentum $$p$$) using the standard Fourier integral $$\int \frac{d^n k}{(2\pi)^n} e^{ikx} |k^2|^s = \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \frac{1}{|x|^{2s+n}}.$$ The $$m=0$$ case is good enough to get the $$\beta$$ function and the wavefunction renormalization $$Z$$ as these do not depend on $$m$$ anyway. If the $$m$$ dependence is important to you you will need to evaluate the Fourier trasnform of the cube of a Bessel "K" function

I do not wish to $$\LaTeX$$ out all the intermediate steps for these integrations, but if you like I can edit my answer to tell you each step along the way. Having said that, let's get down to business.

## $$I_1(p)$$ integration

I will split $$I_1$$ into 2 parts.

$$I_1^a(p) = p^2\int_{0}^1 dx \int_0^{1-x} dy \frac{x y (-x-y+1)}{(x y+(-x-y+1) (x+y))^3}$$

and

$$I_1^b(p) = m^2\int_{0}^1 dx \int_0^{1-x} dy \frac{1}{(x y+(-x-y+1) (x+y))^2}$$

As you can see, I have performed the initial integral over $$z$$, but have strictly followed the definitions provided.

This integral can be done. It can be shown (after much algebra and substitutions) that

$$I_a^1(p) =p^2 \int_0^1 dx\ \frac{3 x}{(x-1) (3 x+1)^2}-\frac{1}{(x-1) (3 x+1)^2} -\frac{8 x \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{(x-1) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}$$

which, I'll admit is a lot less messy than I thought it'd be. The integration at this point is actually somewhat standard. At any rate, it is found after integrating over $$x$$ that

\begin{align*} &I_a^1(p)= p^2 \int_0^1 dx\ \frac{3 x}{(x-1) (3 x+1)^2}-\frac{1}{(x-1) (3 x+1)^2} -\frac{8 x \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{(x-1) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}\\ &=p^2\bigg{(} \frac{\sqrt{-3 x^2+2 x+1} \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{4 (1-x)}+\frac{\sqrt{-3 x^2+2 x+1} \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{3 (3 x+1)^2} + \cdots\\ &-\frac{7 \sqrt{-3 x^2+2 x+1} \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{12 (3 x+1)}-\frac{1}{3 (3 x+1)}\bigg{)} \bigg\rvert_0^1\\ & \implies \boxed{I_1^a(p)= \frac{p^2}{2}} \end{align*}

This integral had no right to simplify so nicely, but sometimes the math gods smile upon us.

Now, we work ok $$I_1^b(p)$$. After performing the $$z$$ integration we have that

$$I_1^b(p) = m^2\int_0^1 dx\ \frac{8 \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}+\frac{2}{-3 x^3+2 x^2+x}$$

Although this looks simpler than the other one $$I_1^a$$ there are divergences lurking. Namely on the endpoints. Therefore, I will use a small parameter $$\epsilon$$ in an attempt to tame it. We arrive at

\begin{align*} &I_1^b(p, \epsilon) = m^2\int_\epsilon^{1-\epsilon}dx \ \frac{8 \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}+\frac{2}{-3 x^3+2 x^2+x}\\ &= 2 m^2\left(\frac{(1-3 \epsilon ) \tanh ^{-1}\left(\frac{1-\epsilon }{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}\right)}{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}+ \underbrace{\log (1-\epsilon )}_{\to 0 \ as\ \epsilon \to 0}-\log (\epsilon )- \underbrace{\frac{(3 \epsilon -2) \tanh ^{-1}\left(\frac{\epsilon }{\sqrt{\epsilon (4- 3 \epsilon )}}\right)}{\sqrt{\epsilon (4-3 \epsilon)}}}_{\to \frac{1}{2}, \ as\ \epsilon \to 0}\right) \end{align*}

The divergences live in the other two terms. So we may substitute in the limits when $$\epsilon \to 0$$ since that's what we will eventually take. We arrive at

\begin{align*} &I_1^b(p, \epsilon) = m^2\int_\epsilon^{1-\epsilon}dx \ \frac{8 \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}+\frac{2}{-3 x^3+2 x^2+x}\\ & \boxed{I_1^b(p, \epsilon) = 2m^2\frac{(1-3 \epsilon ) \tanh ^{-1}\left(\frac{1-\epsilon }{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}\right)}{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}- 2m^2\log (\epsilon )- m^2} \end{align*}

I request that OP tries to expand this in small $$\epsilon$$ before I do, as I still have another integral to do

## $$I_2(p)$$ integration

I must admit, this one pushed back quite a bit. Nevertheless, onward we must go.

The integral to overcome is

$$I_2(p) = \int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right).$$

We split it into two integrals

$$I_2^a = \frac{1}{2} \int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\alpha \beta\right)$$

and

$$I_2^b = -\int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\gamma p^2 +m^2\right)$$

so that our desired integral is the sum.

We know that we can find the indefinite integral of the function multiplying the log, so we close our eyes and hope for the best.

$$I_2^a$$ integration

Integrating over $$z$$ we get that

$$I_2^a = \frac{1}{2}\int_0^1dx \int_0^{1-x} dy \frac{m^2}{(x y+(-x-y+1) (x+y))^2}+\frac{p^2 (x y (-x-y+1))}{(x y+(-x-y+1) (x+y))^3}\times\\ \log\left(x y+(-x-y+1) (x+y) \right)$$

We integrate by parts in $$y$$; taking $$u$$ as the log and $$dv$$ as the coefficient on the log. Then, we have that

$$v= m^2 \left(-\frac{-x-2 y+1}{\left(-3 x^2+2 x+1\right) \left((1-x) y+(1-x) x-y^2\right)}-\frac{4 \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}\right)+p^2 \left(\frac{x^2 (-x-2 y+1)}{2 (3 x+1) \left((1-x) y+(1-x) x-y^2\right)^2}-\frac{4 x \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{(1-x) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}-\frac{x (-x-2 y+1)}{(1-x) (3 x+1)^2 \left((1-x) y+(1-x) x-y^2\right)}\right)$$

and

$$du = dy \frac{x+2 y-1}{x^2+x (y-1)+(y-1) y}$$

Our surface terms reduce to

$$uv\bigg\rvert^{1-x}_0 = \frac{2 \log ((1-x) x) \left((3 x-1) \sqrt{-3 x^2+2 x+1}-8 x \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)\right)}{(x-1) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}$$

And our integral left to be done is

$$\int_0^{1-x} v \ du = \int_0^{1-x} dy\ \Bigg{\{} m^2\left(-\frac{-x-2 y+1}{\left(-3 x^2+2 x+1\right) \left((1-x) y+(1-x) x-y^2\right)}-\frac{4 \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}\right)+p^2 \left(\frac{x^2 (-x-2 y+1)}{2 (3 x+1) \left((1-x) y+(1-x) x-y^2\right)^2}-\frac{4 x \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{(1-x) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}-\frac{x (-x-2 y+1)}{(1-x) (3 x+1)^2 \left((1-x) y+(1-x) x-y^2\right)}\right)\Bigg{\}} \frac{x+2 y-1}{x^2+x (y-1)+(y-1) y}$$

Believe it or not this can be done, but not entirely with elementary functions. Putting the surface terms back and simplifying we have that

\begin{align*} &2I_2^a(p) = \int_0^1 dx\\ &\frac{1}{2 (1-x)^{3/2} (3 x+1)^{5/2}}\bigg{\{} 8 \left(m^2 (3 x+1)+p^2 x\right)+\frac{\sqrt{x (2-3 x)+1} \left(4 m^2 (3 x+1)-p^2 (x-3) x\right)}{x}\\ &+16 \tanh ^{-1}\left(\sqrt{\frac{1-x}{3 x+1}}\right) \bigg{(}\log (4)-\log \left[\left(\sqrt{\frac{1-x}{3 x+1}}+1\right) \left(\frac{x-1}{\sqrt{x (2-3 x)+1}}+1\right)\right]\times\\ & \left(m^2 (3 x+1)+p^2 x\right)-2 m^2 (3 x+1)+p^2 (x-2) x\bigg{)}\\ &-4 \log (x(1-x)) \left((3 x-1) \sqrt{x (2-3 x)+1}-8 x \tanh ^{-1}\left(\sqrt{\frac{1-x}{3 x+1}}\right)\right)\\ &+ \text{Li}_2\left(-\frac{x+\sqrt{1-x} \sqrt{3 x+1}+1}{2 x}\right)-\text{Li}_2\left(-\frac{x-\sqrt{1-x} \sqrt{3 x+1}+1}{2 x}\right) \bigg{\}} \end{align*}

I imagine there are divergences lurking in here, but you can use numerical integration to check. Do your best to do $$I_2^b$$ and get back to me.

I think that's enough for one day.

• Thank you so much for your detailed explanation. I will try to follow these derivation. Commented Jan 2, 2019 at 9:12