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.
Posts
.jpg)