1. introduction

In his book Topics in Analytic Number Theory [?], Rademacher gave a partial fraction decomposition of the partition generating function \(\prod _{j\geq 1}(1-x^j)^{-1}\). He conjectured that the decomposition of the generating function of partitions into at most \(N\) parts, \[ \prod _{j=1}^N \frac{1}{1-x^j} = \sum _{k=1}^N \sum _{\substack{0\leq h<k \\ \gcd (h,k)=1}} \sum _{l=1}^{\lfloor N/k \rfloor } \frac{C_{h,k,l}(N)}{(x-e^{2\pi i h/k})^l}, \] is consistent with it in the sense that the coefficients \(C_{h,k,l}(N)\) converge as \(N\to \infty \) to the coefficients of the decomposition of the unrestricted generating function. Despite attracting the attention of several authors [?, ?, ?], the conjecture has been open since 1973. See Sills and Zeilberger [?] for some further historical remarks. The latter paper presents a recurrence for \(C_{0,1,l}(N)\); the values computed by it do not seem to show convergence, but rather oscillating and unbounded behavior. It is well known, though, that there are number-theoretic problems where the true asymptotics are numerically visible only for very large values. See, e.g., [?] for an example. The present note rigorously confirms the main observation from [?], i.e., we disprove Rademacher’s conjecture.1 To formulate our main result, recall the definition of the dilogarithm function: \(\mathrm{Li}_2(w) = \sum _{k\geq 1}w^k/k^2\), \(|w|<1\). Define \(z_0\approx -1.61 + 7.42i\) as the solution of \begin{equation} \label{eq:z0} \log (1-e^z) + \frac{1}{z}(\mathrm{Li}_2(e^z)-\pi ^2/6) = 0. \end{equation} (It is easy to show that there is a unique root within, say, distance \(1\) of the numerical value given above.) Furthermore, define \(\rho =\exp (i a)\), where \begin{equation} \label{eq:a} a = \frac{ \pi }{2} - \frac 12 \arg \left ( \frac{e^{z_0}}{z_0(1-e^{z_0})} \right ) \approx 1.79. \end{equation}

Note that the number under the first radical sign is real and positive. The period \(p\) of the oscillations is roughly \(32\), as observed by Sills and Zeilberger [?]. It is independent of \(l\), as is the exponential growth order \(b^N\). Moreover, Sills and Zeilberger found that the successive peaks seem to grow exponentially with a factor around \(8\). The (asymptotically) true factor is \(b^p\approx 8.81\). Figures 1 and 2 illustrate the quality of the approximation, which seems to be better for \(l=1\) than for \(l=2\).

In principle, it should be possible to extend our approach from \((h,k)=(0,1)\) to general \(h,k\). Moreover, a natural conjecture is that the period \(2\pi /|\arg (1-e^{z_0})|\) of \(H_l\) is a transcendental number. While there is some literature on transcendence of polylogarithm values (see, e.g., [?]), we are not aware of any result that would imply this.

The rest of the paper is organized as follows. In Section 2, we appeal to the Cauchy integral representation of \(C_{0,1,l}(N)\) and find an asymptotic approximation for its integrand. The new integrand is analysed in Section 3 by the saddle point method. Section 4 completes the proof of Theorem 1 by adding estimates in regions where the asymptotic approximation for the integrand has to be modified or is invalid. In the conclusion, we comment on the error term in (??), and on possible future work.

2. Mellin Transform asymptotics

Since the \(C_{0,1,l}(N)\) are the Laurent coefficients of \(\prod _{j=1}^N (1-x^j)^{-1}\) at \(x=1\), we can express them by Cauchy’s formula: \begin{align} C_{0,1,l}(N) &= \frac{1}{2i\pi } \oint x^{l-1} \prod _{j=1}^N \frac{1}{1-(x+1)^j}dx \notag \\ & = \frac{(-1)^{l-1}}{N^l} \frac{1}{2i\pi } \oint F(z,N) dz, \label{eq:F} \end{align}

where we have substituted \(x+1=e^{z/N}\), and \begin{equation} \label{eq:F def} F(z,N) := e^{f(z,N)} := e^{z/N}N^{l-1}(1-e^{z/N})^{l-1}\prod _{j=1}^N \frac{1}{1-e^{zj/N}}. \end{equation} We wish to replace the integrand \(F\) by an asymptotic approximation, derived by a Mellin transform approach. We do the analysis for \(\Re z<0\), since the factor \(e^{-Nz/2}\) in \begin{equation} \label{eq:refl} \prod _{j=1}^N \frac 1{1-e^{zj/N}} = (-1)^N e^{-z(N+1)/2} \prod _{j=1}^N \frac 1{1-e^{-zj/N}} \end{equation} suggests that the contribution of the left half-circle dominates the integral (??); a rigorous argument is given in Section 4. To take the Mellin transform of \(f=\log F\) w.r.t. \(N\), we have to interpolate between integral values of \(N\). We therefore rewrite the logarithm of the product \(\prod _{j=1}^N\) in (??) as follows: \begin{align*} g(z,N) &:= \log \prod _{j=1}^n \frac{1}{1-e^{zj/N}} \\ &= - \sum _{j=1}^N \log (1-e^{zj/N}) \\ &= \sum _{j=1}^N \sum _{k=1}^\infty \frac{1}{k}e^{zjk/N} = \sum _{k=1}^\infty \frac{1}{k} \frac{1-e^{kz}}{e^{-kz/N}-1}. \end{align*}

Now we can compute the Mellin transform of \(g\) w.r.t. \(N\), for \(\Re (s)<-1\): \begin{align} \mathcal{M}g(z,\cdot )(s)&=\int _0^\infty g(z,x)x^{s-1}dx \notag \\ &= \sum _{k=1}^\infty \frac{1-e^{kz}}{k} \int _0^\infty \frac{x^{s-1}}{e^{-kz/x}-1} dx \notag \\ &= \sum _{k=1}^\infty \frac{1-e^{kz}}{k} (-kz)^s \Gamma (-s)\zeta (-s) \notag \\ &= (-z)^s \Gamma (-s)\zeta (-s) \left (\sum _{k=1}^\infty k^{s-1} - \sum _{k=1}^\infty k^{s-1} e^{kz} \right ) \notag \\ &= (-z)^s \Gamma (-s)\zeta (-s) \Big (\zeta (1-s) - \mathrm{Li}_{1-s}(e^z)\Big ). \label{eq:mellin} \end{align}

Recall that the polylogarithm is defined for \(|w|<1\) and \(\nu \in \mathbb{C}\) by \(\mathrm{Li}_\nu (w) = \sum _{k\geq 1}w^k/k^\nu \). For the integral evaluation used in the third equality, see Titchmarsh [?], p. 18; it already appears in Riemann’s original memoir [?]. By the Mellin inversion formula [?], \(g\) can be recovered by \begin{equation} \label{eq:g mell} g(z,N) = \frac{1}{2i\pi } \int _{-3/2-i\infty }^{-3/2+i \infty } \mathcal{M}g(z,\cdot )(s) N^{-s}ds. \end{equation} We now move the integration line to the right and collect residues. To estimate the resulting integral (and justify Mellin inversion), we first establish a bound on \( \mathrm{Li}_{1-s}(e^z)\) for \(|\Im s|\) large. Note that Pickard [?] studied asymptotics of \(\mathrm{Li}_\nu (w)\) for \(\nu \to 0\) and \(\nu \to \infty \), and wrote that “little is known about behavior in the \(\nu \)-plane except along and near the line \((0,\infty )\).”

We can now find the asymptotics of \(g\) by shifting the integration in (??) to the right, where \(\Re s =8/7\) turns out to be a suitable choice. The polylogarithm \(\mathrm{Li}_{1-s}(e^z)\) is an entire function of \(s\). Moreover, \(\zeta (-s)\) has a simple pole at \(s=-1\), and \(\Gamma (-s)\) has simple poles at the non-negative integers. Because of the factor \(\zeta (1-s)\), the transform (??) has a double pole at \(s=0\), which results in a logarithmic term in the asymptotics of \(g\).

Lemma 3 suggests the approximate integral representation \begin{multline} \label{eq:new int} C_{0,1,l}(N) \\ \approx \frac{(-1)^{l-1}}{N^{l+1/2}(2\pi )^{3/2}i} \int _{|z|=5, \Re z\leq 0} \frac{(-z)^{l-1/2}}{\sqrt{1-e^z}} \exp \left ( \frac{z}{N} + \frac{N}{z} \left ( \mathrm{Li}_2(e^z) - \frac{\pi ^2}{6}\right ) \right ) dz, \end{multline} where \(h\) from (??) has been replaced by zero, except the term \(z/N\), which was retained for better accuracy. Recall that the right half-circle is negligible, as mentioned above and proved in Section (4). Even for small \(N\), the fit is very good for \(l=1\); see Figure 3.

3. Saddle point asymptotics

We now proceed by a saddle point analysis of the integral (??), using the approximation of the integrand provided by Lemma 3. According to this lemma, the factor \( \exp \left ( \frac 1z \left ( \mathrm{Li}_2(e^z) - \frac{\pi ^2}{6}\right ) N \right )\) dominates the integrand in (??). Equating its derivative to zero, we obtain the saddle point \(z_0\) defined in (??). The argument of its axis is (see [?]) \begin{align*} a &= \frac{ \pi }{2} - \frac 12 \arg \frac{d^2}{dz^2} \left .\left ( \frac 1z \left ( \mathrm{Li}_2(e^z) - \frac{\pi ^2}{6}\right )\right ) \right |_{z=z_0} \\ &= \frac{ \pi }{2} - \frac 12 \arg \left ( \frac{e^{z_0}}{z_0(1-e^{z_0})} \right ) \approx 1.79, \end{align*}

and \(\rho =e^{ia}\) is thus the direction of steepest decent. By symmetry, the conjugate \(\bar{z}_0\) is a saddle point, too, and its direction of steepest descent is \(\bar{\rho }\). We now deform the integration

circle as follows (see Figure 4): In the right half-plane, we stay with a half-circle, of radius \(5\). In the left half-plane, we connect the point \(z_1:=5i\) with the point \( z_0-\rho \) by a straight line. We then proceed by a line through the saddle point \(z_0\), up to a point \(z_4(N)\). A vertical line then connects this point to the real axis, to \(z_5(N):=-\sqrt{N}\), and so \(z_4\) must be \[ z_4(N) := -\sqrt{N} + i (\Im z_0-(\sqrt{N}+\Re z_0) \Im \rho / \Re \rho . \] Around the saddle point, we identify a range of width \(O(N^{-39/112})\), delimited by the points \[ z_2(N):=z_0-\rho N^{-39/112} \quad \text{and} \quad z_3(N):=z_0+\rho N^{-39/112}. \] In the lower half-plane, the contour is defined symmetrically. We refer to the line from \(z_2\) to \(z_3\) to the (upper) central part of the contour, as it gives the dominant contribution to the integral (in the upper half-plane). Note that \(-39/112\approx -0.348\) is just a little bit smaller than \(-1/3\). To make the third-order term \(N(z-z_0)^3\) in the local expansion of the integrand negligible, we must have \(z-z_0 \ll N^{-1/3}\). It is convenient to make the central part as large as possible, though, because this causes faster decrease of \(F=e^f\) at \(z=z_2(N)\) and \(z=z_3(N)\), which in turn makes it easier to beat the estimate for \(h\) from Lemma 3. (For details, see the tail estimate in Lemma 4 below.)

Part (i) of Lemma 3 provides the local expansion in the central range, where \(z=z_0+t\rho \), \(-N^{-39/112}\leq t \leq N^{-39/112}\): \begin{multline} \label{eq:loc expans} f(z,N) = - N \log (1 -e^{z_0}) -\tfrac 12 \alpha N t^2 - \tfrac 12 \log N \\ + (l-\tfrac 12)\log (-z_0) -\tfrac 12 (\log 2\pi +\log (1-e^{z_0})) + O(N^{-5/112}). \end{multline} (Note that the expansion was simplified by using the defining equation (??) of \(z_0\).) The constant \[ \alpha := -\frac{\rho ^2e^{z_0}}{z_0(1-e^{z_0})} \approx 0.028 \] is real and positive. Since \[ \int _{-N^{-39/112}}^{N^{-39/112}}\exp (-\tfrac 12 \alpha N t^2)dt \sim \frac{1}{\sqrt{\alpha N}}\int _{-\infty }^{\infty }e^{-x^2/2}dx = \sqrt{\frac{2\pi }{\alpha N}}, \] with exponentially decaying error term, the saddle point integral has the asymptotics \begin{equation} \label{eq:sp int} \int _{z_2(N)}^{z_3(N)} e^{f(z,N)}dz = \frac{\rho (-z_0)^{l-1/2}}{ \sqrt{\alpha (1-e^{z_0})}} \frac 1N (1-e^{z_0})^{-N}(1+O(N^{-5/112})). \end{equation} The integral over the lower saddle point segment is \[ \int _{\bar{z}_3(N)}^{\bar{z}_2(N)}e^{f(z,N)}dz = - \overline{\int _{z_2(N)}^{z_3(N)} e^{f(z,N)}dz}, \] and thus the contribution of both saddle points to the integral (??) is \[ \frac{(-1)^{l-1}}{\pi N^l} \Im \left ( \int _{z_2(N)}^{z_3(N)} e^{f(z,N)}dz \right ). \] By (??), we see that this gives the right hand side of (??).

To show that the two small line segments containing the saddle points \(z_0\) resp. \(\bar{z}_0\) capture the asymptotics of the full integral (??), we have to show that the remaining part of the contour in Figure 4 is negligible. By conjugation, it clearly suffices to consider the upper half-plane. We begin with the part that, additionally, lies in the half-plane \(\Re z \leq N^{-7/8}\). In the next section, we show that the integral over the remaining part tends exponentially to zero.

4. Estimates close to the imaginary axis and in the right half-plane

In the preceding section, have gave an asymptotic evaluation of the integral (??), where the contour was deformed as in Figure 4, and \(\Re z < -N^{-7/8}\). We now show that the remaining part of the contour is negligible. Close to the imaginary axis, where \(-N^{-7/8} \leq \Re z \leq 0\), we are outside of the validity region of the Mellin transform asymptotics of Lemma 3. We thus estimate the integrand in (??) directly.

Finally, we estimate the integral over the right half-circle in (??), which completes the proof of Theorem 1. By the reflection formula (??), we can recycle part of the analysis from the left half-plane.

5. conclusion

The error term that we obtained in (??) can be improved a bit by considering more terms of the local expansion (??) of \(f\) in the saddle point analysis. Getting the correct order of the error term, i.e., the next term in the asymptotic expansion of \(C_{0,1,l}(N)\), needs some work, though. As only the first term on the right-hand side of (??) was used to define the saddle point \(z_0\), the logarithmic terms in (??) contribute a non-vanishing first order term \(O(z-z_0)=O(N^{-39/112})\) to the expansion (??). To improve it, we need to replace \(z_0\) by a better approximation of the actual saddle point of the whole integrand \(F=e^f\). But then, the tail estimate in Lemma 4 becomes more involved, because not only the width, but also the location of the saddle point segment depends on \(N\).

Perhaps more importantly, we comment on possible future work. Recall that Rademacher’s conjecture essentially says that the operations of limit and partial fraction decomposition commute in the present setting. While our result refutes the conjecture, it does not clarify the relation between the partial fraction decompositions of \(\prod _{j\geq 1}(1-x^j)^{-1}\) and \(\prod _{j=1}^N(1-x^j)^{-1}\); it would be surprising if there was none at all. O’Sullivan [?] suspects that some modified version of the conjecture might hold, and presents numerical evidence for convergence of \(C_{h,k,l}(N)\) in terms of Cesàro means.