Closed form for ∑∞n=1ψ(n+54)(1+2n)(1+4n)2\sum_{n=1}^\infty\frac{\psi(n+\frac{5}{4})}{(1+2n)(1+4n)^2}

This question came up in the process of finding solution to another problem. Eventually, the problem was solved avoiding calculation of this sum, but it looks quite interesting on its own. Is there a closed form for
n=1ψ(n+54)(1+2n)(1+4n)2,
where ψ(z)=Γ(z)Γ(z) is the digamma function?

Answer

Yes, there is!

The derivation, which involves a lot of computer-assisted manipulations, is a little bit long, so I won’t put it here just now, but the answer is
2CCγ12Cπ+14γππ26γπ28+1164π36Li3(1+i2)3Clog2+12γlog2+πlog2+32(log2)2+316π(log2)2+74ζ(3)ψ(54),
where C is the Catalan constant, γ is Euler’s gamma, and Li3 is a polylogarithm, and (edited) the last term ψ(5/4)=4γπ2log8 is there because everything below computes the sum over n0, so I need to subtract n=0 from the answer below to get the sum over n1.

Edit to explain the calculation.
The way to derive this is to write the sum as
18S(54|14,14)18S(54|14,12),
where I’ve introduced the notation
S(β|α1,α2,)=n0ψ(n+β)(n+α1)(n+α2).

The intermediate sums can be found in closed form mainly using the integral representation of the digamma function, in the form
ψ(z1)ψ(z2)=10dt1t(tz21tz11),
a sum representation in the form
ψ(z+1)+γ=k11k1k+z,
and using also the definition of the Lerch transcendent and its relation to the hypergeometric function and the incomplete beta function:
Φ(z,s,a)=n0zn(n+a)s,
Φ(z,1,a)=zaB(z,a,0)=1aF(1,a;1+a;z).

The intermediate sum S(54;14,14) can be found using
n0ψ(n+54)+γ(n+14)2=n0k11(n+14)2(1k1k+n+14)=k1ψ(k+14)ψ(14)k2=4F(1,1,1,12,2,54)+10logtlog(1t)t(1t)3/4dt,
where
Li2(1)Li2(t)=Li2(1t)+logtlog(1t),
so
S(54|14,14)=γψ1(1/4)+n0ψ(n+54)+γ(n+14)2.

Using the above relationship between incomplete beta function, the fact that this hypergeometric is a repeated integral of the incomplete beta function (which you know because of the ones on top and matching twos on the bottom) and the fact that there is a nice closed form for the incomplete beta function with a rational first parameter and zero second parameter, like so:
B(z,p/q,0)=0l<qe2πilp/qlog(1z1/qe2πil/q),p,qZ,
it can be simplified to
F(1,1,1,12,2,54)=732π312Li3(1+i2)+32π(log2)2+π2log814ζ(3).

The second intermediate sum can be simplified using partial fractions on 1(n+14)(n+12), and the fact that
n01(n+α1)(n+α2)=ψ(α1)ψ(α2)α1α2.
Then
S(54|14,12)+γψ(12)ψ(14)1214=n0ψ(n+54)+γ(n+14)(n+12)=k1ψ(k+14)ψ(12)k(k14),
where I've expanded difference of digamma functions as an infinite sum over k and performed the sum over n. This expression can be handled using the integral representation for ψ mentioned above, giving
10dt1t((12log22π)t12t34(4log(1t)+163tF(1,34;74;t)))=16C+43π28πlog212(log2)3.

Putting everything together, and using Mathematica to evaluate the easier sums, gives the expression I gave above.

Basically, the most important steps in the derivation are the integral representation for ψ(x)ψ(y) and the explicit formula for B(z,β,0) when β is a rational number. By luck all the integrals then simplify to something Mathematica can do in closed form.

I don't know any standard reference for this; the whole question is basically just computing
ddϵ|ϵ=0F(1,12,14,14,54+ϵ32,54,54,54),
and in general derivatives of hypergeometric functions are given by Kampe de Feriet functions, which don't have a closed form in terms of hypergeometric functions. So in general there should be no closed form, but in this case the parameters are just right that all integrals can be done.

When Q(n) is some rational function having a partial fraction expansion kqk(n+αk)s, the sum
n0Q(n)(ψ(n+β)ψ(β))
can be expressed in terms of sums
n0ψ(n+β)ψ(β)(n+α)s=10B(v,β,0)Φ(v,s,α)dv,
where Φ is the Lerch transcendent's derivative. If the sum on the left diverges, it is necessary to consider the asymptotic expansion
n0ψ(n+β)ψ(β)n+αzn=Θ(log(1z))2+Θ(log(1z))+S(β,α)+o(1),z1
and only take the non-divergent term S(β,α) as the value of the sum at z1 (the divergent terms should be independent of α and cancel out when the partial fractions of Q(n) are added together).

So the problem of evaluating the sum
n0Q(n)(ψ(n+β)ψ(β))
reduces to calculating asymptotic expansions of integrals of the form
z10B(u,β,0)Φ(zu,s,α)du,z1.

Attribution
Source : Link , Question Author : Laila Podlesny , Answer Author : Kirill

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