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Using two different approaches to evaluating the dimensionally regularized ($d=4-2\epsilon$ dimensional Euclidean space), equal mass ($x=m^2$), 2-loop vacuum Feynman diagram $$ \begin{align} I(x) &= \int\frac{\mathrm{d}^dp}{\pi^{d/2}}\frac{\mathrm{d}^dk}{\pi^{d/2}} \frac1{(k^2+x)(p^2+x)((k+p)^2+x)} \\\ &= \int _0^{\infty }\int _0^{\infty }\int _0^{\infty } \frac{e^{-x(s_1+s_2+s_3)}}{\left(s_1s_2+s_2s_3+s_3s_1\right)^{d/2}} \mathrm{d}s_1\mathrm{d}s_2\mathrm{d}s_3 \\\ &= x^{d-3}\,\Gamma\left(2-\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right) \,f(d)\,, \end{align} $$ the following hypergeometric identity arises $$ \begin{align} f(d) &=f_1(d) = 2\, {}_2F_1\left(1,\frac{d-1}{2};\frac{3}{2};-\frac{1}{3}\right) -4^{2-d} 3^{(d - 3)/2} B\left(\frac{3 - d}2, \frac{3 - d}2\right) \\ &= f_2(d) = \frac{4}{3} \left( {}_2F_1\left(1,\frac{d-1}{2};\frac{3}{2};-\frac{1}{3}\right)+\frac{1 }{d-3} {}_2F_1\left(1,\frac{d-1}{2};\frac{5-d}{2};-\frac{1}{3}\right) \right)\,, \end{align} $$ where the second term in $f_1(d)$ can be reduced with the identity $B(x,x)=2^{1-2x}B(x,1/2)$.

The identity $f_1(d)=f_2(d)$ checks out numerically and (provided no mistakes have been made in the calculations) it should be identically true. So far I have been unable to find a direct proof of the identity.

Can anyone here prove this identity or suggest a good approach? A computer proof (using human checkable code/steps) is acceptable.


For convenience I've provided the Mathematica InputForm of the two functions:

f1[d_] := 2 Hypergeometric2F1[1, (d - 1)/2, 3/2, -1/3] - 
          2^(4 - 2 d) 3^((d - 3)/2) Beta[(3 - d)/2, (3 - d)/2]

f2[d_] := 4/3 (Hypergeometric2F1[1, (d - 1)/2, 3/2, -1/3] + 
          1/(d - 3) Hypergeometric2F1[1, (d - 1)/2, (5 - d)/2, -1/3])

Aside:
$f_1(d)$ comes from direct integration using Feynman parameters (my own calculation, I don't know of a reference that includes it).
$f_2(d)$ comes from direct integration using the Mellin-Barnes representation (the result presented above is a version of eqn(33) of hep-ph/9304303, see also references within) .


Edit: I just noticed this MO answer that mentions the HolonomicFunctions package for Mathematica. It shows that both functions obey the recursion
$(4+4 d)f_i(d+4)+(4-7 d)f_i(d+2)+(-6+3 d)f_i(d)=0$,
but of course, the integral diverges for integer $d\geq4$ and I need to prove the relation for all $d<4$ (dimensional reduction) or for all complex $d$ (dimensional regularization).


Aside #2: Maybe I've been viewing this problem backwards, and I should not be using hypergeometric identities to check the Feynman integrals, but rather using the Feynman integrals as inspiration for new hypergeometric identities. See the new paper: Finding new relationships between hypergeometric functions by evaluating Feynman integrals

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    $\begingroup$ Standard methods for algorithmically verifying hypergeometric identities are described in Petkovsek, Wilf, and Zeilberger's A=B: math.upenn.edu/~wilf/AeqB.html $\endgroup$ Commented May 30, 2011 at 12:45
  • $\begingroup$ @Qiaochu It's been a long time since I flicked through A=B... I'll have to go back and look again. $\endgroup$
    – Simon
    Commented May 30, 2011 at 13:13

1 Answer 1

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Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctions by Christoph Koutschan.

In your particular example,

Annihilator[f1[d], {S[d]}]

shows that this function satisfies the recurrence \begin{equation} (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. \end{equation}

Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:

(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify

After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$

But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...

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  • $\begingroup$ Thanks! I like the idea of using Carlson's Theorem to show that $f_1-f_2$ is identically zero. It might be a little tricky though, since both $f_i(d)$ have poles at odd $d>2$. But I guess that knowing that $f_1(d)-f_2(d)=0$ for even integers is enough to use Carlson's theorem - so now I just need to check the growth conditions. $\endgroup$
    – Simon
    Commented May 31, 2011 at 0:36
  • $\begingroup$ @Armin: Ok, I haven't really done this kind of analysis before and am not sure how to get bounds on the expansion of $f_1-f_2$ around $d=\infty$... I tried using the integral form of the hypergeometric to extract some suitable asymptotics, but got nowhere. Would you mind expanding your answer on the use of Carlson's theorem? $\endgroup$
    – Simon
    Commented Jun 1, 2011 at 1:39

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