Analytic Continuation of the Lerch Transcendent

We define the following important Dirichlet series as the Lerch transcendent,

Φ(z,s,a) := sum_[n=0 to +infinity] (z^n)/(n+a)^s , a ∉ {0}∪Z^- ; s ∈ C for |z|<1 and Re(s)>1 when |z|=1

Note that z=1 gives the Hurwitz zeta function, and a=0 gives the s-logarithm of z as notable special cases (and with a=1 this is just the Riemann zeta function).

In the case when z=exp (2πiλ) we have the Lerch (or periodic) zeta function,

L(λ,s,a) := sum_[n=0 to +infinity] (e^(2πiλn))/(n+a)^s

We have the following integral representation for Φ,

Φ(z,s,a)Γ(s) = sum_[n=0 to +infinity] (z^n) integral_[0 to +infinity] (t^(s-1)).e^(-(n+a)t) dt , for Re(a)>0, Re(s)>0

with the change of variable t := (n+a)t in the Euler integral representation for the Gamma function.

We now interchange sum and integral (which is justified here) to get,

Φ(z,s,a)Γ(s) = integral_[0 to +infinity] (t^(s-1)).e^(-at) sum_[n=0 to +infinity] (z.e^(-t))^n dt

The sum within the integral is a geometric series, and on summing we have our identity,

Φ(z,s,a)Γ(s) = integral_[0 to +infinity] [(t^(s-1)).e^(-at) / (1 - z.e^(-t))] dt

for Re(a)>0; |z|≤1, z≠1, Re(s)>0; z=1, Re(s)>1

This suggest a contour integral of the form,

(2πi).I(z,s,a) := integral_[H] (t^(s-1)).e^(at) / (1 - z.e^t) dt

Where H is a Hankel contour where,

H= C_1 ∪ C_2 ∪ C_3

and C_1 is the portion of H which travels in a straight line from +infinity just under the positive real axis, which then connects to C_2 which traverses a semi-circle of radius ρ≥0 which again connects to C_3 which returns to +infinity just above the positive real axis.

So with C_1 parametrised by t=r.e^(-πi) , C_2 parametrised by t=ρ.e^(iθ) and C_3 parametrised by t=r.e^(πi) , we have,

(2πi).I(z,s,a) = integral_[+infinity to ρ] [[r^(s-1).e^(-πis).e^(πi).e^(-πi).e^(-ra)] / (1 - z.e^(-r))] dr +

integral_[-π to +π] [[ρ^(s-1).e^(sθi).e^(-θi).e^(aρ.e^(iθ)).ρi.e^(iθ)] / (1 - z.e^(ρ.e^(iθ)))] dθ +

integral_[ρ to +infinity] [[r^(s-1).e^(πis).e^(-πi).e^(πi).e^(-ra)] / (1 - z.e^(-r))] dr

On simplifying,

(2πi).I(z,s,a) = (e^(πis) - e^(-πis)).integral_[ρ to +infinity] [r^(s-1).e^(-ra) / (1 - z.e^(-r))] dr + i(ρ^s). integral_[-π to +π] [e^(iθs + aρ.e^(iθ)) / (1 - z.e^(ρe^(iθ)))] dθ

The last integral in θ tends to 0 as ρ->0 (in the region to which we continue to, *details to be supplied).

We then have,

lim_[ρ->0] I(z,s,a) = (sin (πs) / π) Φ(z,s,a)Γ(s)

Recalling the formula Γ(s)Γ(1-s) = (π / sin (πs)) gives the contour integral representation,

Φ(z,s,a) = [Γ(1-s) / (2πi)] integral_[H] [t^(s-1).e^(at) / (1 - z.e^t)] dt ------- (1)

Re(a)>0, |arg(-t)| =< π

cf.

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1103052188

http://abel.math.harvard.edu/~knill/kam/papers/denjoy.pdf

A 'Natural Proof' of the Gauss Multiplication formula II: Theorem and Proof

We are now ready to prove the result:

Theorem (Gauss's Multiplication formula for Γ)

Γ(Nt) = (2π)^((1-N)/2) . N^(Nt - 1/2) . prod_[k=0 to N-1] Γ(t + k/N) ------- (2)

Proof:

From (1) we have that,

√(2π) / Γ(Nt) = prodz_[m=0 to +infinity] (m + Nt) = prod_[k=0 to N-1] prodz_[m=0 to +infinity] ((mN + k) + Nt)

it is easy to see that there is a bijection between the product indices on either side of the last equality since m=0 on the right gives the terms of the product on the left from m=0 up to m=N-1 and so on.

Then,

√(2π) / Γ(Nt) = prod_[k=0 to N-1] prodz_[m=0 to +infinity] N.(m + (t + k/N))

From the lemma we established in the last post we have,

√(2π) / Γ(Nt) = prod_[k=0 to N-1] N^[Z(0, t + k/N)] prodz_[m=0 to +infinity] (m + (t + k/N))

We then recognise the regularised product from (1) again and along with the value of Z(0, t + k/N), have,

√(2π) / Γ(Nt) = prod_[k=0 to N-1] N^[1/2 - (t + k/N)] . √(2π) / Γ(t + k/N) = N^(1/2 - Nt) . (2π)^(N/2) . prod_[k=0 to N-1] 1/Γ(t + k/N)

Which gives the required formula on rearrangement.

QED.

The proof flows naturally from the regularized determinant expression for the Gamma function unlike standard proofs which match up zeros and poles of either side of (2), which is why the term 'natural' was used to describe it. We also note that the simple functional equation Γ(z+1) = z.Γ(z) is obvious from (1).

I am much indebted to Professors Kurokawa and Wakayama for their wonderful paper.

For more on regularization see Basic Analysis Of Regularized Series And Products by Jay Jorgenson and Serge Lang.

A 'Natural Proof' of the Gauss Multiplication formula I: Preliminaries

The following proof of the Gauss Multiplication formula for the Gamma function is an elaboration of the original that appears in the excellent paper 'Zeta Regularizations' by N Kurokawa and M Wakayama in Acta Applicandae Mathematicae Volume 81, Number 1 / March, 2004. We follow their exposition throughout.
http://www.springerlink.com/content/q185862551668217/

It is by far the most natural proof I have come across of this theorem even though the fundamental idea is from the theory of regularized determinants.

A few preliminaries

The notation introduced here will be used throughout.
Observe that for a given Hurwitz zeta function, Z_λ (s, t) = sum_[λ_n] (λ_n + t)^(-s),
∂[Z_λ (0, t)]/∂s = - sum_[λ_n] log (λ_n + t)

So we define the zeta-regularized product for the sequence (λ_n)_{n=0,...},
prodz_[λ_n] (λ_n + t) := exp (- ∂[Z_λ (0, t)]/∂s)

For convenience when λ_n = n we shall denote the resulting classical Hurwitz zeta function by Z.

The main result that will be employed is Lerch's formula,

∂[Z(0, t)]/∂s = -(1/2).log (2π) + log Γ(t)

where log is the natural logarithm and Γ is the Gamma function as defined at http://en.wikipedia.org/wiki/Gamma_function

Note that Z'(0) = -(1/2).log (2π) where Z is the Riemann zeta function.

For a proof the reader can refer to http://ocw.nctu.edu.tw/upload/fourier/supplement/Zeta-Function.pdf or the chapter on the Zeta and Gamma functions in S Lang's Complex Analysis (4th ed.), Springer Graduate Texts in Mathematics.

We now have the immediate corollary that,

prodz_[n=0 to +infinity] (n + t) = √(2π) / Γ(t) ------- (1)

(await latter half for proof)...