Result of this integral in large Lambda limit

$$\int \text{d}^4 q \, \frac{1}{(q^2 + m^2)\left(1+\frac{q^2}{\Lambda^2}\right)^2} =2 \pi^2\left(\frac{\Lambda^2}{2}-m^2 \log \frac{\Lambda}{m} \right) + o(\Lambda^0)$$

How to get this result? The notation ##o(\Lambda^0)## means all terms constant in Lambda, which we ignore because we are interested in a large ##\Lambda## limit. Also, the implicit region of integration is all of ##\mathbb{R}^4##.

I managed to switch to spherical coordinates and integrate over the angular variables to pull put a factor of the surface area of the unit 3-sphere. The rest of the integral picks up a factor of ##q^3## and becomes an integration over dq from 0 to infinity.

From there, what is a quick way to get to the result? Is there some trick that lets you see the large Lambda behaviour while ignoring the constant terms?

Siupa said:

From there, what is a quick way to get to the result? Is there some trick that lets you see the large Lambda behaviour while ignoring the constant terms?

Explicitly writing out the result of the angular-integration step you describe gives:$$\intop_{\mathbb{\mathbb{R}}^{4}}\text{d}^{4}q\,\frac{1}{(q^{2}+m^{2})\left(1+\frac{q^{2}}{\Lambda^{2}}\right)^{2}}=2\pi^{2}\intop_{0}^{\infty}dq\frac{q^{3}}{(q^{2}+m^{2})\left(1+\frac{q^{2}}{\Lambda^{2}}\right)^{2}}$$Now if you google "Feynman integration tricks" or "Feynman parameterization" you can doubtless find a clever method or three for manually evaluating the semi-infinite integral over ##q##. But as for me, I find it much quicker to open Wolfram Mathematica, feed it the integral (which evaluates in a few seconds) and then power expand around ##\Lambda=\infty##:

renormalize said:

Explicitly writing out the result of the angular-integration step you describe gives:$$\intop_{\mathbb{\mathbb{R}}^{4}}\text{d}^{4}q\,\frac{1}{(q^{2}+m^{2})\left(1+\frac{q^{2}}{\Lambda^{2}}\right)^{2}}=2\pi^{2}\intop_{0}^{\infty}dq\frac{q^{3}}{(q^{2}+m^{2})\left(1+\frac{q^{2}}{\Lambda^{2}}\right)^{2}}$$Now if you google "Feynman integration tricks" or "Feynman parameterization" you can doubtless find a clever method or three for manually evaluating the semi-infinite integral over ##q##. But as for me, I find it much quicker to open Wolfram Mathematica, feed it the integral (which evaluates in a few seconds) and then power expand around ##\Lambda=\infty##:View attachment 331395

Unfortunately Mathematica is only useful if you need to find the result or verify that it's true, but I don't need that. I already know that the result is true, I need to find out how to get there.

I'll try Feynman parameterization

Siupa said:

Unfortunately Mathematica is only useful if you need to find the result or verify that it's true, but I don't need that. I already know that the result is true, I need to find out how to get there.

Can you clarify what motivates your interest in this particular integral? Is it primarily:

Mathematical: you're focused on examining techniques of integration?
or
Physical: this integral arises in a physics problem (e.g., perturbative solutions and renormalization of an interacting quantum field theory)?

If it's the latter, why not make use of available tools, like integral tables or algebra software, so you can concentrate on the physics? (In my opinion, it's really not that much different than, say, using a calculator to sum a long list of integers instead of tediously doing it by hand.)

renormalize said:

Can you clarify what motivates your interest in this particular integral? Is it primarily:

Mathematical: you're focused on examining techniques of integration?
or
Physical: this integral arises in a physics problem (e.g., perturbative solutions and renormalization of an interacting quantum field theory)?

If it's the latter, why not make use of available tools, like integral tables or algebra software, so you can concentrate on the physics? (In my opinion, it's really not that much different than, say, using a calculator to sum a long list of integers instead of tediously doing it by hand.)

I need to prepare for an oral exam for a course in QFT. If my professor asks me to derive the 1-loop renormalization of the 2-point function in phi^4 theory, they expect me to be able to follow through the steps and arrive at a result. I don't think they'll be satisfied if I answer with "let me just pull up Mathematica"

Susbstituting [itex]q = \Lambda x = \Lambda \sqrt{u}[/itex] gives [tex]\begin{split}
\int_0^\infty \frac{q^3}{(q^2 + m^2)\left(1 + \frac{q^2}{\Lambda^2}\right)^2}\,dq &=
\Lambda^2 \int_0^\infty \frac{x^3}{(x^2 + \epsilon^2)(1 + x^2)^2}\,dx \\
&= \frac{\Lambda^2}{2} \int_0^\infty \frac{u}{(u + \epsilon^2)(1 + u)^2}\,du\end{split}[/tex] where [itex]\epsilon = m/\Lambda[/itex]. The integral can be done exactly for any [itex]\epsilon[/itex] by partial fractions.

pasmith said:

Susbstituting [itex]q = \Lambda x = \Lambda \sqrt{u}[/itex] gives [tex]\begin{split}
\int_0^\infty \frac{q^3}{(q^2 + m^2)\left(1 + \frac{q^2}{\Lambda^2}\right)^2}\,dq &=
\Lambda^2 \int_0^\infty \frac{x^3}{(x^2 + \epsilon^2)(1 + x^2)^2}\,dx \\
&= \frac{\Lambda^2}{2} \int_0^\infty \frac{u}{(u + \epsilon^2)(1 + u)^2}\,du\end{split}[/tex] where [itex]\epsilon = m/\Lambda[/itex]. The integral can be done exactly for any [itex]\epsilon[/itex] by partial fractions.

Unfortunately partial fraction decomposition doesn't work: the original integral converges, but after PFD you get two divergent integrals. I'm sure that the divergences "cancel out" to give the same finite result that you would have got without PDF, but to handle the two new individually divergent integrals you need to introduce regulators, but this defeats the purpose because the original integral is already supposed to be regularised (the parameter Lambda is precisely the UV regulator)

Siupa said:

Unfortunately partial fraction decomposition doesn't work: the original integral converges, but after PFD you get two divergent integrals. I'm sure that the divergences "cancel out" to give the same finite result that you would have got without PDF, but to handle the two new individually divergent integrals you need to introduce regulators, but this defeats the purpose because the original integral is already supposed to be regularised (the parameter Lambda is precisely the UV regulator)

We have [tex]
\int_0^\infty \frac{u}{(u + \epsilon^2)(1 + u)^2}\,du = \int_0^\infty \frac{\epsilon^2}{(1 - \epsilon)^2}\left(\frac{1}{1 + u} - \frac{1}{u + \epsilon^2}\right) + \frac{1}{1 - \epsilon^2}\frac{1}{(1 + u)^2}\,du.[/tex] There is in fact no difficulty with [tex]\begin{split}
\int_0^\infty \frac{1}{u + 1} - \frac{1}{u + \epsilon^2} \,du &= \lim_{R \to \infty} \int_0^R \frac{1}{1 + u} - \frac{1}{u + \epsilon^2}\,du \\
&= \lim_{R \to \infty} \log \left(\frac{1 + R}{\epsilon^2 + R}\right) - \log(1/\epsilon^2) \\
&= 0 + 2\log \epsilon \end{split}[/tex] but if you don't like that you can treat the first term as [tex]\begin{split}
\frac{-\epsilon^2}{1 - \epsilon^2}\int_0^\infty \frac{1}{u^2 + (1 + \epsilon^2)u + \epsilon^2}\,du
&= \frac{-2\epsilon^2}{(1 - \epsilon^2)^2} \int_{v_0}^\infty \frac{1}{\sinh v}\,dv \\
&= \frac{2\epsilon^2}{(1 - \epsilon^2)^2} \ln \tanh(v_0/2)
\end{split}[/tex] where [tex]
\cosh v_0 = \frac{1 + \epsilon^2}{1 - \epsilon^2}[/tex] so that [tex]\tanh v_0 = \frac{2\epsilon}{1 + \epsilon^2} = \frac{2\tanh (v_0/2)}{1 + \tanh^2(v_0/2)}.[/tex]

EDIT: Corrected sign in denominator of [itex]\tanh v_0[/itex].

Siupa said:

I need to prepare for an oral exam for a course in QFT. If my professor asks me to derive the 1-loop renormalization of the 2-point function in phi^4 theory, they expect me to be able to follow through the steps and arrive at a result. I don't think they'll be satisfied if I answer with "let me just pull up Mathematica"

Ok, so this is akin to a homework problem. Perhaps you should post your question in the homework forum and show your attempt at a solution?

pasmith said:

We have [tex]
\int_0^\infty \frac{u}{(u + \epsilon^2)(1 + u)^2}\,du = \int_0^\infty \frac{\epsilon^2}{(1 - \epsilon)^2}\left(\frac{1}{1 + u} - \frac{1}{u + \epsilon^2}\right) + \frac{1}{1 - \epsilon^2}\frac{1}{(1 + u)^2}\,du.[/tex] There is in fact no difficulty with [tex]\begin{split}
\int_0^\infty \frac{1}{u + 1} - \frac{1}{u + \epsilon^2} \,du &= \lim_{R \to \infty} \int_0^R \frac{1}{1 + u} - \frac{1}{u + \epsilon^2}\,du \\
&= \lim_{R \to \infty} \log \left(\frac{1 + R}{\epsilon^2 + R}\right) - \log(1/\epsilon^2) \\
&= 0 + 2\log \epsilon \end{split}[/tex] but if you don't like that you can treat the first term as [tex]\begin{split}
\frac{-\epsilon^2}{1 - \epsilon^2}\int_0^\infty \frac{1}{u^2 + (1 + \epsilon^2)u + \epsilon^2}\,du
&= \frac{-2\epsilon^2}{(1 - \epsilon^2)^2} \int_{v_0}^\infty \frac{1}{\sinh v}\,dv \\
&= \frac{2\epsilon^2}{(1 - \epsilon^2)^2} \ln \tanh(v_0/2)
\end{split}[/tex] where [tex]
\cosh v_0 = \frac{1 + \epsilon^2}{1 - \epsilon^2}[/tex] so that [tex]\tanh v_0 = \frac{2\epsilon}{1 - \epsilon^2} = \frac{2\tanh (v_0/2)}{1 + \tanh^2(v_0/2)}.[/tex]

Thank you, yes, that actually works if one takes care with putting an upper cutoff and then taking the limit afterwards. I will try the other method that the other user proposed with Feynman parameters, but in any case this works. Thank you!

renormalize said:

Ok, so this is akin to a homework problem. Perhaps you should post your question in the homework forum and show your attempt at a solution?

I don't think that this falls under the definition of "homework", but maybe I didn't read the description of each forum well enough, if that's the case, apologies.

As for showing my attempt at a solution, I think I did, I described my first steps with changing variables and then integrating the angular part. If I knew the next step after that was, I would have also already known the solution, so I don't know why what I showed wasn't enough to meet the requirements for "attempting a solution".

Anyways the other user solved it using partial fraction decomposition, and actually after that I tried your initial suggestion with Feynman parameter and that also worked, so it's fine. Thank you

To continue my earlier post, from [tex]
\cosh v_0 = \frac{1 + \epsilon^2}{1 - \epsilon^2}[/tex] we can obtain [tex]\begin{split}
\epsilon^2 &= \frac{\cosh v_0 - 1}{\cosh v_0 + 1} \\
&= \frac{2\sinh^2(v_0/2)}{2\cosh^2(v_0/2)} \\
&= \tanh^2(v_0/2)\end{split}[/tex] so that [tex]
\frac{2\epsilon^2}{(1 - \epsilon^2)^2}\ln \tanh(v_0/2) = \frac{2\epsilon^2 \ln \epsilon}{(1 - \epsilon^2)^2}.[/tex]

A very nice treatment of how to evaluate such integrals can be found in

P. Ramond, Field Theory: A Modern Primer, Addison-Wesley, Redwood City, Calif., 2 edn. (1989).