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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

4 added new reference

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.
Finding new relationships between hypergeometric functions by evaluating Feynman integrals

3 small change to make recursion clearer (and make my question active again)

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
$(4+4 d)S_d^4+(4-7 d)S_d^2+(-6+3 d)=0$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).

2 added note about HolonomicFunctions package
1