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 formal summation formula

Define the Bernoulli numbers B_n by,

t / (e^t - 1) = sum_[n=0 to +infinity] (B_n).t^n / n!

Let ∂ f(t) = d[f(t)]/dt and put ∂^(-1) f(t) = integral f(t) dt

Note that,

exp [n∂ f(t)] = sum_[k=0 to +infinity] (n^k).∂^(n) f(t) / k! = f(t+n)

where the last inequality is by taking Taylor's theorem as a formal identity.

Then,

sum_[n=0 to +infinity] f(t+n) = sum_[n=0 to +infinity] exp [n∂ f(t)] = (1 - e^∂)^(-1) f(t) , by summing the geometric series.

Then,

sum_[n=0 to +infinity] f(t+n) = -sum_[n=0 to +infinity] (B_n)(∂^(n-1) f(t)) / n!

Let t->0 and we have,

sum_[n=0 to +infinity] f(n) = -sum_[n=0 to +infinity] (B_n)(∂^(n-1) f(0)) / n! -------- (1)

Put f(n) = n^k for k ∈ Z+ and we have all terms vanishing on the right hand side except that at n=k+1

For n=0 on the left there is no term so the left hand side is just the Riemann zeta function at -k, and we have,

Z(-k) = -(B_(k+1)).k! / (k+1)! = -(B_(k+1)) / (k+1)

We define the Bernoulli polynomials B_n (x) = sum_[k=0 to n] (nCk).(B_k).x^(n-k)

These have generating function,

t.e^(xt) / (e^t - 1) = sum_[n=0 to +infinity] (B_n (x)).t^n / n!

Put f(n) = (n+μ)^k for k ∈ Z+ and μ a constant in (1) and we have the Hurwitz zeta function Z on the left hand side and,

Z(k, μ) = -sum_[r=0 to k+1] (B_r)(k!).μ^(k-r+1) / (r!)(k-r+1)!

then,

Z(k, μ) = -(1/(k+1)) sum_[r=0 to k+1] ((k+1)Cr).(B_r).μ^(k-r+1) = -(B_(k+1) (μ)) / (k+1)

Which is an identity we needed in our proof of the Gauss multiplication theorem for the Gamma function.

We also remark that a two-variable generalisation is possible by considering f(t+y) rather than f(t) and deriving,

sum_[k=0 to n-2] f(t+y+k) = [sum_[k=0 to n-1] e^((y+k)D)] f(t) = [e^(yD) / (e^D - 1)] (f(t+n-1) - f(t))

(by summing the geometric series as before).

Whence by using the definition of the Bernoulli polynomials the operator in square brackets can be expanded as an infinite series. Putting f(t) = log (t) we can obtain an asymptotic expansion for log Γ(n+y) in powers of the reciprocal of n, which is Stirling's approximation (for y=0) with the constant involved expressed in series form!

Our formal manipulations can be shown to match up with the analytic extensions of these functions.

See G H Hardy, Divergent Series, 2nd ed. AMS Chelsea, 1991 (cf. sec.13.12) for more on the Euler-Maclaurin summation formula.

The Polylogarithm

Define the k-logarithm, L_k by,

L_k (z) = sum_[n=1 to +infinity] z^n / n^k , for |z|<1

We will derive an interesting identity for this function which illustrates the discrete Fourier transform.

Let ζ be an N th root of unity with ζ not =1 (i.e- ζ is primitive). Consider the sum,

N^(k-1) sum_[ζ^N = 1] L_k (ζz)

Using the definition of L_k (z) we can write this as,

N^(k-1) sum_[n=1 to +infinity] (z^n / n^k) sum_[ζ^N = 1] ζ^n

Note that, sum_[i=0 to N-1] ζ^i = (1 - ζ^N) / (1 - ζ) = 0

by summing the geometric progression. Now note that this is satisfied also by ζ^2, ... , ζ^(N-1) so these are the other roots of unity apart from 1 and ζ (at ζ=N we again get 1 and cycle through). But this is again the same geometric series and so,

sum_[ζ^N = 1] ζ^n = 0 , for all n>1 with n not =N.

Then in our sum,

N^(k-1) sum_[ζ^N = 1] L_k (ζz)

N^(k-1) sum_[lN

(N^(k-1)).z^(nN) / [(n^k).(N^(k-1))] = z^(nN) / n^k for n=1,...

Then,

N^(k-1) sum_[ζ^N = 1] L_k (ζz) = L_k (z^N)

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 II: A Lemma

We continue with our treatment of the multiplication formula for the Gamma function following Kurokawa and Wakayama.

We will require the following:

Lemma

prodz_[n ∈ J] μ.(a_n + b) = [μ^Z(0,b)].prodz_[n ∈ J] (a_n + b)

for μ, b constants.

Z_[μ.a_n] (s, t) = sum_[n ∈ J] (μ.a_n + t)^(-s) = [μ^(-s)].[Z_[a_n] (s, t/μ)] , by taking μ^(-s) out of the sum and noting the resulting zeta function.

Then,

Z_[μ.a_n] '(s, t) = -(μ^(-s)).(log μ).[Z_[a_n] (s, t/μ)] + (μ^(-s)).[Z_[a_n] '(s, t/μ)] , by differention using the product rule.

With s=0,

-Z_[μ.a_n] (0, t) = log [μ^(Z_[a_n] (0, t/μ))] - Z_[a_n] '(0, t/μ)

Which gives,

exp (-Z_[μ.a_n] '(0, t)) = [μ^(Z_[a_n] (0, t/μ))]. exp (-Z_[a_n] '(0, t/μ))

t=μb gives the desired result.

We note that, in the classical case,

Z(-n, b) = -B_[n+1] (b) / (n+1) , where B_n (x) is the n th Bernoulli polynomial on x, for n a natural number and b a constant. Then, Z(0, b) = (1/2) - b

(a divergent series based proof of this will be given in a later post)

We finish the proof with the coming post!

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)...