Hi Jason, All you need is the relation In[270]:= FunctionExpand[%266, x > 0] Out[270]= BarnesG[x] == E^(-(1/2) (-1 + x) x - PolyGamma[-2, x]) (2 \[Pi])^(x/2) Gamma[x]^(-1 + x) Then In[271]:= Table[ FunctionExpand[%], {x, {1/2, 1/3, 1/4, 1/6, 2/3, 3/4, 5/6}}] Out[271]= {True, BarnesG[1/3] == ( 3^(1/72) E^( 1/9 + \[Pi]/(18 Sqrt[3]) - PolyGamma[1, 1/3]/(12 Sqrt[3] \[Pi])))/( Glaisher^(4/3) Gamma[1/3]^(2/3)), BarnesG[1/4] == E^(3/32 - Catalan/(4 \[Pi]))/( Glaisher^(9/8) Gamma[1/4]^(3/4)), BarnesG[1/6] == E^( 5/72 + \[Pi]/(12 Sqrt[3]) - PolyGamma[1, 1/3]/(8 Sqrt[3] \[Pi]))/( 2^(1/72) 3^(1/144) (Glaisher Gamma[1/6])^(5/6)), BarnesG[2/3] == ( 3^(1/72) E^( 1/9 - \[Pi]/(18 Sqrt[3]) + PolyGamma[1, 1/3]/(12 Sqrt[3] \[Pi])))/( Glaisher^(4/3) Gamma[2/3]^(1/3)), BarnesG[3/4] == E^(3/32 + Catalan/(4 \[Pi]))/( Glaisher^(9/8) Gamma[3/4]^(1/4)), BarnesG[5/6] == E^( 5/72 - \[Pi]/(12 Sqrt[3]) + PolyGamma[1, 1/3]/(8 Sqrt[3] \[Pi]))/( 2^(1/72) 3^(1/144) Glaisher^(5/6) Gamma[5/6]^(1/6))} In[272]:= N[%] Out[272]= {True, True, True, True, True, True, True} --Bill Gosper On Mon, Apr 14, 2014 at 10:51 AM, Wolfram Technical Support < support@wolfram.com> wrote:
Hello Bill Gosper,
Thank you for the further information.
I have passed those notes along to our development team. I will let you know if I hear anything back about this issue.
We are always interested in improving Mathematica and I want to thank you again for taking the time to contact us about this issue. If you find any further issues, please feel free to contact us again.
Sincerely,
Jason Grigsby Wolfram Technology Group Wolfram Research, Inc www.wolfram.com
On: 04/13/2014 10:56pm billgosper@gmail.com submitted the following Case: On Tue, Mar 11, 2014 at 2:53 AM, Bill Gosper wrote:
Papers of mine and your own Victor Adamchik give Integrate[Log[Gamma[t]],{t,0,r}] for various rational r. We also give this integral in terms of BarnesG. But FunctionExpand[BarnesG[r]] only "works" for r=n/2, which it gives in terms of the goddam useless Glaisher symbol. Zeta'[-1] spontaneously turns into 1/12 - Log[Glaisher]. Why?? Why not then arbitrarily smash every Zeta[3] to be Apery? Or Apéry, to make it harder to type.
Here are G[1/3] and -1/3 with z1 for Zeta[-1], to avoid the Glaisher braindamage.
This should be Zeta'[-1], but the following are wrong!
Out[67]= {BarnesG[1/3] -> 3^(25/72)* E^((1/72)*(96*z1 + (Sqrt[3]*(-PolyGamma[1, 1/3] + PolyGamma[1, 2/3]))/Pi))* Gamma[1/3]^(1/Gamma[4/3])*Gamma[4/3]^(4/3), BarnesG[-(1/3)] -> 3^(1/72)* E^((1/72)*(96*z1 + (Sqrt[3]*(PolyGamma[1, 1/3] - PolyGamma[1, 2/3]))/Pi))* Gamma[-(1/3)]^(1/Gamma[2/3])* Gamma[2/3]^(2/3)}
Profuse apologies to all for inadequate testing! These should read
{BarnesG[1/3]==E^(1/72 (8+(Sqrt[3] (-PolyGamma[1,1/3]+PolyGamma[1,2/3]))/\[Pi]))/(3^(47/72) (E^(1/12-(Zeta^\[Prime])[-1]))^(4/3) ((1/3)!)^(2/3)), BarnesG[-(1/3)]==-((3 3^(1/72) E^(1/72 (8+(Sqrt[3] (PolyGamma[1,1/3]-PolyGamma[1,2/3]))/\[Pi])) ((1/3)!/\[Pi])^(4/3))/(2 2^(1/3) (E^(1/12-(Zeta^\[Prime])[-1]))^(4/3)))}
My (inadequate) excuse: FunctionExpand just flat out returned a wrong answer! Nonreproducible, of course, but I have the .nb recording the event. My guess is that I damaged Mathematica somehow while trying to banish the vexatious Glaisher symbol.
Numerical test: In[175]:= N[ReleaseHold[%]]
Out[175]= {True, False}
HAH! So much for testing. They're both true, dammit. (There's an invisible HoldForm around the Zeta'[-1]s) --rwg
In[68]:= Zeta'[-1]
Out[68]= 1/12 - Log[Glaisher]
FunctionExpand also appears to disunderstand the tuplication formula Product[Hyperfactorial[-(i/n) + z], {i, 0, -1 + n}] == (E^(z1/n - n*z1)*Hyperfactorial[n*z]^(1/n))/ n^((1 + 6*n*z + 6*n^2*z^2)/(12*n))
whose need it artificially creates for itself when given the Alternating Sign Matrix formula In[171]:= FunctionExpand[Product[(3 k + 1)!/(n + k)!, {k, 0, n - 1}], n > 0 && n \[Element] Integers]
Out[171]= (3^(2 - 5 n + 3/2 n (1 + n)) 8^(-1 + n) (BarnesG[5/3] BarnesG[7/3])^(-2 + n) (BarnesG[8/3] BarnesG[10/3])^(1 - n) BarnesG[2/3 + n] BarnesG[1 + n] BarnesG[4/3 + n] BarnesG[2 + n]^( 2 - n) BarnesG[3 + n]^(-1 + n) Gamma[2 + n]^( 1 - n))/(BarnesG[1 + 2 n] Gamma[1 + n])
In[175]:= FunctionExpand[Table[%171, {n, 5}]]
Out[175]= {1, 2, ( 9 BarnesG[5/3] BarnesG[7/3] BarnesG[11/3] BarnesG[13/3])/( 5 BarnesG[8/3]^2 BarnesG[10/3]^2), ( 2187 BarnesG[5/3]^2 BarnesG[7/3]^2 BarnesG[14/3] BarnesG[16/3])/( 7000 BarnesG[8/3]^3 BarnesG[10/3]^3), ( 1594323 BarnesG[5/3]^3 BarnesG[7/3]^3 BarnesG[17/3] BarnesG[19/3])/( 274400000 BarnesG[8/3]^4 BarnesG[10/3]^4)}
This is an integer sequence! (A005130, 1,2,7,42,...). --Bill Gosper In[71]:= $LicenseID
Out[71]= "L3290-7570"