# Analytic continuation of the Dirichlet generating series of the multiplicative partition function

Apologies for the lengthy question, but it seems it's the only way i can convey my thoughts. Consider the Dirichlet series: $$\kappa(s)=\prod_{m=2}^{\infty}\frac{1}{1-m^{-s}}=\sum_{n=1}^{\infty}\frac{\rho(n)}{n^{s}}\;\;\;\;(\Re(s)>1)$$ Where $\;\rho(n)$ is the number of different ways to represent the integer $n$ as a product of powers of distinct integers, each greater than one. We say that two representations are equivalent if they differ only in the ordering of multiplications. Now consider : $$\log\kappa(s)=-\sum_{n=2}^{\infty}\log(1-n^{-s})=\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{n^{-ks}}{k}=\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k}$$ The function $\log\kappa(s)$ has an accumulation of poles at the reciprocal of evey +ive integer, an essential singularity at $s=0$, and a natural boundary at $\Re(s)=0$.However, using formal arguments, i was able to do the following:

We denote by $\left \lfloor x \right \rfloor$ the floor function, and define the function $\Omega(x)$: $$\Omega(x)=\sum_{n=1}^{\infty}\frac{\left \lfloor x^{1/n} \right \rfloor-1}{n}$$ And we have: $$\log\kappa(s)=s\int_{1}^{\infty}\Omega(x)x^{-s-1}dx$$
Using the Fourier expansion of the sawtooth function, we have the formal representation : $$\log\kappa(s)=\sum_{n=1}^{\infty}\frac{3-ns}{2n(ns-1)}+s\int_{1}^{\infty}\frac{\sin(2\pi nz)}{\pi n z(z^{s}-1)}dz$$ Now we consider the formal negation of $s$. By taking the difference : $$\log\kappa(s)-\log\kappa(-s)=\sum_{n=1}^{\infty}\frac{2s}{(ns)^{2}-1}-s\int_{1}^{\infty}\frac{\sin(2\pi nz )}{\pi n z}dz$$ and making use of the partial fraction expansion of $\pi\cot(\pi s)$: $$\pi\cot(\pi s)=\frac{1}{s}+2s\sum_{n=1}^{\infty}\frac{1}{s^{2}-n^{2}}$$ We may write: $$\log\kappa(s)-\log\kappa(-s)=\log(2\pi)^{s/2}-\pi\cot\left(\frac{\pi}{s}\right)$$The constant factor $\log(2\pi)^{1/2}$ originates from: $$\sum_{n=1}^{\infty}\int_{1}^{\infty}\frac{\sin(2\pi n z)}{\pi n z}dz=\int_{1}^{\infty}\frac{1}{z}\left(\left \lfloor z \right \rfloor-z+\frac{1}{2}\right)dz=1+\zeta^{'}(0)=1-\log(2\pi)^{1/2}$$

We could prove our last result using the following definition of the Riemann zeta function for $\Re(s)>-1$: $$\zeta(s)=\frac{1}{2}+\frac{1}{s-1}+\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}+\frac{1}{2}-\frac{1}{x}\right)x^{s-1}e^{-x}dx$$ or: $$\frac{\zeta(ns)-1}{n}=\frac{3-ns}{2n(ns-1)}+\frac{s}{\Gamma(1+ns)}\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}+\frac{1}{2}-\frac{1}{x}\right)x^{sn-1}e^{-x}dx$$ taking the summation over $n\geqslant 1$, one has: $$\log\kappa(s)=\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{3-ns}{2n(ns-1)}+s\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}+\frac{1}{2}-\frac{1}{x}\right)\frac{E^{N}_{s,1}(x^{s})-1}{x}e^{-x}dx$$

Where $E_{s,1}^{\infty}$ is the Mittag-Leffler function, which admits the beautiful continuation : $$E_{-s,1}^{\infty}(x^{-s})=1-E_{s,1}^{\infty}(x^{s})$$

Formally,we take the difference $\log\kappa(s)-\log\kappa(-s)$ to obtain: $$\log\kappa(s)-\log\kappa(-s)=s-\pi\cot\left(\frac{\pi}{s}\right)-s\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}+\frac{1}{2}-\frac{1}{x}\right)\frac{e^{-x}}{x}dx$$ The rightmost integral is finite, and has the numerical value: $$\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}+\frac{1}{2}-\frac{1}{x}\right)\frac{e^{-x}}{x}dx=1+\zeta^{'}(0)=1-\log(2\pi)^{1/2}$$ Thus: $$\log\kappa(s)-\log\kappa(-s)=\log(2\pi)^{s/2}-\pi\cot\left(\frac{\pi}{s}\right)$$ We use the partial fraction expansion: $$\frac{1}{z^{s}-1}=\sum_{k=-\infty}^{\infty}\frac{1}{s\log z+2\pi i k}-\frac{1}{2}$$ Thus, we write the formal representation : $$\log\kappa(s)=\sum_{n=1}^{\infty}\frac{3-ns}{2n(ns-1)}+\int_{1}^{\infty}\left(\frac{1}{2}-z+\left \lfloor z \right \rfloor \right )\left(\sum_{k=-\infty}^{\infty}\frac{1}{z\left(\log z+\frac{2\pi i k}{s}\right)}-\frac{s}{2z}\right)dz$$ The leftmost summation is divergent due to the presence of the harmonic series, and so is the integral term corresponding to $k=0$. However, if we split the summation into convergent and divergent parts, the divergence is absorbed by the integral counterpart into a constant $\lambda$. Namely: $$\lambda=\lim_{N\rightarrow \infty}\lim_{\alpha \rightarrow 1}\int_{\alpha}^{\infty}\left(\frac{1}{2}-z+\left \lfloor z \right \rfloor \right )\frac{dz}{z\log z}-\frac{1}{2}\sum_{n=1}^{N}\frac{1}{n}$$ The remaining part of the summation is convergent, and is given by: $$\sum_{n=1}^{\infty}\frac{1}{n(ns-1)}=-\gamma-\psi\left(\frac{s-1}{s}\right)$$ Where $\gamma$ is the Euler-Mascheroni constant, and $\psi(s)$ is the digamma function. The rightmost term of the integral is just a constant times $s$: $$s\int_{1}^{\infty}\left(\frac{1}{2}-z+\left \lfloor z \right \rfloor \right )\frac{dz}{2z}=\frac{1+\zeta^{'}(0)}{2}s=\frac{1-\log(2\pi)^{1/2}}{2}s$$ And we're left with: $$\int_{1}^{\infty}\sum_{k=1}^{\infty}\left(\frac{1}{2}-z+\left \lfloor z \right \rfloor \right )\left(\frac{1}{z\left(\log z+\frac{2\pi i k}{s}\right)}+\frac{1}{z\left(\log z-\frac{2\pi i k}{s}\right)}\right)dz$$ This expression is invariant under the reflection $s\rightarrow -s$, and in conjunction with the recurrence formula of the digamma function, could be used to prove our result. If we denote by $\Psi(s)$ the integral above, then we may write: $$\log\kappa(s)=\lambda-\gamma-\psi\left(\frac{s-1}{s}\right)-\frac{1-\log(2\pi)^{1/2}}{2}s+\Psi(s)$$ EDIT :

Using the definition of the Riemann zeta function for $\Re(s)>-1$, It is easily seen that if $s$ is kept real, then the following holds:

$$\int_{1}^{\infty}\left(\frac{1}{2}-z+\left \lfloor z \right \rfloor \right )\left(\frac{1}{z\left(\log z\pm\frac{2\pi i k}{s}\right)}\right)dz=\int_{0}^{\infty}\left(\frac{\zeta(y)}{y}+\frac{1}{2y}-\frac{1}{y-1}\right)e^{\mp\frac{2\pi ik}{s}y}dy$$

We integrate the RHS by parts, and take the summation of the $+k$ and $-k$ terms over $k\geqslant 1$ to obtain:

$$\Psi(s)=\frac{s}{2\pi i }\int_{0}^{\infty}\frac{d}{dy}\left(\frac{\zeta(y)}{y}+\frac{1}{2y}-\frac{1}{y-1} \right )\log\left(-e^{-\frac{2\pi iy}{s}}\right)dy$$ My questions :

1) This is all formal, and defies the observation that $\log\kappa(s)$ has a natural boundary. But, can we make it rigorous, and interpret the line $\Re(s)=0$ as a branch cut, as opposed to a natural boundary?

2) The digamma term induced me to consider the related function : $$\tilde{\kappa}(s)=e^{\gamma+\psi\left(\frac{s-1}{s} \right )}\kappa(s)$$. This function has the advantage of 'singularity cancellation'. In addition, its $\log$ has no accumulation of poles. This made me believe that there is hope, even via 'generalized analytic continuation' - not in the Weierstraß sense - .

-
You say that the function has "an essential singularity at $s=0$, and thus has a natural boundary at $\Re(s)=0$". What do you mean by the "and thus"? For example, the function $e^{-1/s}$ has an essential singularity at $s=0$, but no line is a natural boundary. – Greg Martin Mar 4 '13 at 23:27
yes, you're right ... i should've said : it has an accumulation of singularities. – mohammad-83 Mar 5 '13 at 6:46
Is it obvious/known that $s\mapsto\log\kappa(s)$ has a natural boundary at ${\Re(s)=0}$ ? Obviously the form you first give is not defined after this line, but is it clear that no analytic continuation can be performed near any point of the line? – Loïc Teyssier Mar 5 '13 at 8:48
Estermann proved the following theorem : Let $c$ be any real number, $\beta<c-\frac{1}{2}$, and $\alpha_{n}=n^{c}+O(n^{\beta})$, then $$\sum_{n=1}^{\infty}\alpha_{n}\left(\zeta(ns)-1\right)$$ is singular at every point of the imaginary axis . – mohammad-83 Mar 5 '13 at 10:28
Hi Mohammad. I think one should avoid doing minor spelling corrections because your question pops up at the main page. To increase interest in your question, I suggest you either put a bounty on it or ask a more general question about (generalized) analytic continuation. Btw, I voted up in the beginning. I find the question very interesting:) Best, Marc – Marc Palm Mar 7 '13 at 12:48