Sho nuff, Mma doesn't know this class of MeijerG special values: {F[1, c] -> (1/4)*(-2 + Cos[c]*(-4*CosIntegral[c] + c*(Pi - 2*SinIntegral[c])) + 2*Sin[c]*(Pi + c*CosIntegral[c] - 2*SinIntegral[c])), F[2, c] -> (1/16)*(-12 + 2*CosIntegral[c]*((-8 + c^2)*Cos[c] + 7*c*Sin[c]) + 7*c*Cos[c]*(Pi - 2*SinIntegral[c]) - (-8 + c^2)*Sin[c]*(Pi - 2*SinIntegral[c])), F[3, c] -> (1/96)*(2*(-44 + c^2) + 6*(-16 + 5*c^2)*Cos[c]*CosIntegral[c] - 2*c*(-57 + c^2)*CosIntegral[c]*Sin[c] - c*(-57 + c^2)*Cos[c]*(Pi - 2*SinIntegral[c]) - 3*(-16 + 5*c^2)*Sin[c]*(Pi - 2*SinIntegral[c]))} (These are sufficient to initialize the 3rd order recurrence, below.) As in Warren's paper, F[n,c] is "just" a MeijerG/(2 n! sqrt(π)): {F[0, c] -> (1/2)*(-2*Cos[c]*CosIntegral[c] + Sin[c]*(Pi - 2*SinIntegral[c])), F[1, c] -> MeijerG[{{-1}, {}}, {{0, 0, 1/2}, {}}, c^2/4]/(2*Sqrt[Pi]), F[2, c] -> MeijerG[{{-2}, {}}, {{0, 0, 1/2}, {}}, c^2/4]/(4*Sqrt[Pi]), F[3, c] -> MeijerG[{{-3}, {}}, {{0, 0, 1/2}, {}}, c^2/4]/(12*Sqrt[Pi])} F[n,c] is a polynomial of first degree in Sin[c], Cos[c], SinIntegral[c], CosIntegral[c], and π, and (for n>1) degree 2*Floor[n/2] in c itself, with a messy profusion of crossterms. Maybe we could do better with a finite Abel-Plana using the function log(n). E.g., Out[128]= Log[m!] == -m - 2*Integrate[ArcTan[(m*y)/(1 + m + y^2)]/ (-1 + E^(2*Pi*y)), {y, 0, Infinity}] + (1/2 + m)*Log[1 + m] In[129]:= % /. m -> -1/2 Out[129]= Log[Pi]/2 == 1/2 - 2*Integrate[-(ArcTan[y/(2*(1/2 + y^2))]/(-1 + E^(2*Pi*y))), {y, 0, Infinity}] In[130]:= N[List @@ %] Out[130]= {0.572365, 0.572365} But I guess that just brings us pretty well back to Binet. The argument of the atan is <1 for the whole 0<y<oo, so integrating the usual atan powerseries produces nice convergence, but the termwise integrals are mysterious, even for m=1. Peachy source of definite integral identities, though: Log[(1/4)!] == (1/4)*(-1 - 8*Integrate[ArcTan[y/(5 + 4*y^2)]/(-1 + E^(2*Pi*y)), {y, 0, Infinity}] + 3*Log[5/4]) On Fri, Dec 30, 2011 at 1:57 PM, Bill Gosper <billgosper@gmail.com> wrote:
wds>many formulas about gamma function are in my paper:http://rangevoting.org/WarrenSmithPages/homepage/gammprox.pdf
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Interesting stuff. Random remarks: P4 middle is garbled, at least in my Preview. Did you mean asymptosy vs asymptopia?
P6, In[16]:= F[0, c] == Assuming[c > 0,Integrate[
x^(2*n + 1)/(1 + x^2)^(n + 1)*E^(-c*x) /. n -> 0, {x, 0, \[Infinity]}]]
Out[16]= F[0,c] == (-Cos[c])*CosIntegral[c] + (1/2)*Sin[c]*(Pi - 2*SinIntegral[c])
In[12]:= G[0, c] == Assuming[c > 0, Integrate[
x^(2*n)/(1 + x^2)^(n + 1)*E^(-c*x) /. n -> 0, {x, 0, \[Infinity]}]]
Out[12]= G[0, c] == CosIntegral[c]*Sin[c] + (1/2)*Cos[c]*(Pi - 2*SinIntegral[c])
In[14]:= H[0, c] == Assuming[c > 0, Integrate[
x^(2*n)/(1 + x^2)^(n)*E^(-c*x) /. n -> 0, {x, 0, \[Infinity]}]]
Out[14]= H[0, c] == 1/c
Eliminating G and H from the F recurrence, 4*(n-1)*n*F(n)=2*(n-1)*(6*n-5)*F(n-1)-(12*n^2-32*n+c^2+22)*F(n-2)+2*(n-2)*(2*n-3)*F(n-3)
Mma gets only MeijerG for n>0 in all of these.
P7 (40), see http://www.tweedledum.com/rwg/idents.htm near bottom of 1st screen.
(42) and (43) raise the interesting question of evaluating
Product[((1/(12*n)) + 1)*((n - 1/2)*(E^(1/n) - 1))^n,{n, 1, Infinity}]
(i.e., prod(((n-1/2)*(%e^(1/n)-1))^n*(1+1/12/n),n,1,inf)
--"cwg"
Man, it's hard to get even a riesable value. .90058064316485891020 is probably only good to 12 digits. --rwg