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. 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)} 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"