A 2F1 Hypergeometric identity from a Feynman integral - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:21:32Z http://mathoverflow.net/feeds/question/66442 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66442/a-2f1-hypergeometric-identity-from-a-feynman-integral A 2F1 Hypergeometric identity from a Feynman integral Simon 2011-05-30T12:29:45Z 2011-12-01T12:38:26Z <p>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) &amp;= \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)} \\ &amp;= \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 \\ &amp;= 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) &amp;=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) \\ &amp;= 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 $B(x,x)=2^{1-2x}B(x,1/2)$.</p> <p>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.</p> <p>Can anyone here prove this identity or suggest a good approach? A computer proof (using human checkable code/steps) is acceptable.</p> <hr> <p>For convenience I've provided the Mathematica <code>InputForm</code> of the two functions:</p> <pre><code>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]) </code></pre> <hr> <p><strong>Aside:</strong><br> $f_1(d)$ comes from direct integration using Feynman parameters (my own calculation, I don't know of a reference that includes it).<br> $f_2(d)$ comes from direct integration using the Mellin-Barnes representation (the result presented above is a version of eqn(33) of <a href="http://arxiv.org/abs/hep-ph/9304303" rel="nofollow">hep-ph/9304303</a>, see also references within) .</p> <hr> <p><strong>Edit:</strong> I just noticed <a href="http://mathoverflow.net/questions/37787/proving-a-hypergeometric-function-identity/37836#37836" rel="nofollow">this MO answer</a> that mentions the <a href="http://www.risc.jku.at/research/combinat/software/HolonomicFunctions/index.php" rel="nofollow">HolonomicFunctions</a> package for Mathematica. It shows that both functions obey the recursion<br> $(4+4 d)f_i(d+4)+(4-7 d)f_i(d+2)+(-6+3 d)f_i(d)=0$,<br> but of course, the integral diverges for integer $d\geq4$ and I need to prove the relation for all $d&lt;4$ (dimensional reduction) or for all complex $d$ (dimensional regularization).</p> <hr> <p><strong>Aside #2:</strong> 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: <a href="http://arxiv.org/abs/1108.6019" rel="nofollow">Finding new relationships between hypergeometric functions by evaluating Feynman integrals</a></p> http://mathoverflow.net/questions/66442/a-2f1-hypergeometric-identity-from-a-feynman-integral/66449#66449 Answer by Armin Straub for A 2F1 Hypergeometric identity from a Feynman integral Armin Straub 2011-05-30T13:41:59Z 2011-05-30T13:41:59Z <p>Since you are using Mathematica, you definitely want to take a look at the extremely useful package <a href="http://www.risc.jku.at/research/combinat/software/HolonomicFunctions/index.php" rel="nofollow">HolonomicFunctions</a> by Christoph Koutschan.</p> <p>In your particular example,</p> <pre><code>Annihilator[f1[d], {S[d]}] </code></pre> <p>shows that this function satisfies the recurrence $$(4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0.$$</p> <p>Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:</p> <pre><code>(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 </code></pre> <p>After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$</p> <p>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 <a href="http://en.wikipedia.org/wiki/Carlson%27s_theorem" rel="nofollow">Carlson's Theorem</a>. I hope that helps...</p>