# A 2F1 Hypergeometric identity from a Feynman integral

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

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

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

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...

• 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. May 31, 2011 at 0:36
• @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? Jun 1, 2011 at 1:39