proof of Gauss’ digamma theorem
The first formula is the logarithmic derivative of
Γ(x+n)=(x+n-1)(x+n-2)⋯xΓ(x) |
By the partial fraction decomposition satisfied by the ψ function,
ψ(pq)+γ=∞∑n=0(1n+1-qp+nq)=limt→1-∞∑n=0(1n+1-qp+nq)tp+nq |
using Abel’s limit theorem.
Now,
∞∑n=0(1n+1-qp+nq)tp+nq=∞∑n=0tp+nqn+1-∞∑n=0qtp+nqp+nq=tp-q∞∑n=0t(n+1)qn+1-q∞∑n=0tp+nqp+nq |
Since
-ln(1-t)=∞∑n=1tnn |
the first term is
-tp-qln(1-tq) |
Using the algorithm for extracting every qth term of a series (http://planetmath.org/ExtractingEveryNthTermOfASeries), the second term is
q-1∑n=0ω-npln(1-ωnt) |
and therefore
∞∑n=0(1n+1-qp+nq)tp+nq | =-tp-qln(1-tq)+q-1∑n=0ω-npln(1-ωnt) | ||
=-tp-qln1-tq1-t-(tp-1-1)ln(1-t)+q-1∑n=1ω-npln(1-ωnt) |
Let t→1- to get
ψ(pq)=-γ-lnq+q-1∑n=1ω-npln(1-ωn) |
Replace p by q-p and add the two expressions to obtain
ψ(pq)+ψ(q-pq)=-2γ-2lnq+2q-1∑n=1cos(2πnpq)ln(1-ωn) |
The left side is real, so it is equal to the real part of the right side. But
ℜ(ln(1-ωn))=ln|1-ωn|1/2=ln|(1-cos2πnq)2+sin22πnq|1/2=12ln(2-2cos2πnq) |
and so
ψ(pq)+ψ(q-pq)=-2γ-2lnq+q-1∑n=1cos(2πnpq)ln(2-2cos2πnq) | (1) |
But
ψ(x)-ψ(1-x)=ddxln(Γ(x)Γ(1-x))=-πcotπx |
by the Euler reflection formula and thus
ψ(pq)-ψ(q-pq)=-πcotπpq | (2) |
Add equations (1) and (2) to get
ψ(pq) | =-γ-π2cotπpq-lnq+12q-1∑n=1cos2πnpqln(2-2cos2πnq) | ||
=-γ-π2cotπpq-lnq+q-1∑n=1cos2πnpqln(2sinπnq) |
where the last equality holds since
ln(2-2cos(2θ))=ln(2-2(1-2sin2θ)=ln(4sin2θ)=2ln(2sinθ) |
References
- 1 G.E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge University Press, 2001.
-
2
J.L. Jensen [1915-1916], An elementary exposition of the theory of the gamma function
, Ann. Math. 17, 124-166.
Title | proof of Gauss’ digamma theorem |
---|---|
Canonical name | ProofOfGaussDigammaTheorem |
Date of creation | 2013-03-22 16:24:07 |
Last modified on | 2013-03-22 16:24:07 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 5 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 30D30 |
Classification | msc 33B15 |