# Computing the Hardy-Ramanujan asymptotic formula using method of steepest descent/saddle point method

I am trying to obtain and prove the Hardy-Ramanujan asymptotic approximation formula given by: $$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}},$$

by using Dedekind's eta function

$$\eta(z)=e^{\frac{i\pi z}{12}}\prod_{n=1}^{\infty}(1-e^{2\pi inz})=\prod_{n=1}^{\infty}\frac{1}{1-z^{n}}$$

and the method of steepest descent (or saddle point method). I am stuck on formulating the integral to which this method can be applied.

I understand that we can use the generating function: $$f(z)=1+\sum_{n=1}^{\infty}p(n)z^{n}$$

to represent $$p(n)$$ as the integral

$$p(n)=\frac{1}{2\pi i}\int_{C}\frac{f(z)}{z^{n+1}} \, dz$$

around a closed path $$C$$ entirely within the unit circle enclosing the origin. Using Dedekind's eta function gives me that

$$f(z)=e^{\frac{i\pi z}{12}}(\eta(z))^{-1}.$$

Since $$\eta(-\frac{1}{z})=\sqrt{\frac{z}{i}}\eta(z)$$ I have that

$$f(z)=\sqrt{\frac{z}{i}}e^{\frac{i\pi}{12z}}e^{\frac{i\pi z}{12}}f\left(-\frac{1}{z}\right).$$

However, if $$z$$ is restricted appropriately and $$z \to 0$$, then $$\Im(-\frac{1}{z}) \to \infty$$ implies that $$f(-\frac{1}{z}) \to 1$$ since $$f(z)=1+\mathcal{O}(e^{-2\pi y})$$, where $$z=x+iy$$, $$y \geq 1$$.

So now the integral I am interested in is $$p_{1}(n)=\int_{\gamma}\sqrt{\frac{z}{i}}e^{\frac{i\pi}{12z}}e^{\frac{i\pi z}{12}}e^{-2\pi inz} \, dz.$$

Is this the correct integral to consider? How do I now apply the method of steepest descent/saddle point method to obtain the Hardy-Ramanujan asymptotic approximation formula?

• You can combine the generating function of the $p(n)$ with Cauchy's integral formula, to obtain the integral representation to start with.
– Gary
Commented Dec 7, 2020 at 12:58
• @Gary I have made some changes that hopefully incorporated what you have suggested. Commented Dec 8, 2020 at 13:45
• I have got $$p(n) = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {e^{n(x + iy)} f(x + iy)dy} ,\quad f(w) = \prod\limits_{n = 1}^\infty {\frac{1}{{1 - e^{ - nw} }}} .$$ I saw the proof using a saddle point analysis once. The main contribution will come from a neighbourhood of $y=0$ that shrinks with $n$. It is a tedious calculation and at one point you indeed use the functional equation of the eta function.
– Gary
Commented Dec 8, 2020 at 19:51

OP is basically correct. Recall that the Euler function is

$$\varphi(\tau)~:=~\prod_{m\in\mathbb{N}}(1-q^m)=\frac{1}{\sum_{n\in\mathbb{N}_0}p(n)q^n}, \qquad q~\equiv~e^{2\pi i \tau},$$

and that the Dedekind eta function is

$$\eta(\tau)~:=~q^{\frac{1}{24}}\varphi(\tau).$$

We calculate the partition function

\begin{align} p(n)~=~&\oint_{|q|<1}\frac{dq}{2\pi i}\! \frac{q^{-(n+1)}}{\varphi(\tau)}\cr ~=~&\int_{i\delta+\left[-\frac{1}{2},\frac{1}{2}\right]} \! d\tau~ \frac{q^{\frac{1}{24}-n}}{\eta(\tau)}, \qquad \delta~>~0, \cr ~=~&\int_{i\delta+\left[-\frac{1}{2},\frac{1}{2}\right]} \! d\tau~ \sqrt{-i\tau}\frac{q^{\frac{1}{24}-n}}{\eta(\tilde{\tau})}, \qquad \tilde{\tau}~\equiv~-\tau^{-1}, \cr ~=~&P^{\prime}(n),&\end{align}

where

\begin{align} P(n) ~:=~&\int_{i\delta+\left[-\frac{1}{2},\frac{1}{2}\right]} \! \frac{d\tau}{2\pi}~ \frac{q^{\frac{1}{24}-n}\tilde{q}^{-\frac{1}{24}} }{\sqrt{-i\tau}\varphi(\tilde{\tau})}, \qquad \tilde{q}~\equiv~e^{2\pi i \tilde{\tau}},\cr ~=~&\frac{1}{\sqrt{\lambda}}\int_{i\lambda\delta+\left[-\frac{\lambda}{2},\frac{\lambda}{2}\right]} \! \frac{d\bar{\tau}}{2\pi}~ \frac{e^{-\lambda S(\bar{\tau})}}{\sqrt{-i\bar{\tau}\rule[0pt]{0pt}{6pt}}\varphi(-\lambda\bar{\tau}^{-1})}, \qquad \bar{\tau}~\equiv~\lambda\tau, \end{align}

where

$$\lambda~\equiv~\sqrt{n-\frac{1}{24}}~>~0$$

is large, and where

\begin{align} S(\bar{\tau})~:=~&2\pi i\left(\bar{\tau}-\frac{1}{24\bar{\tau}}\right),\cr S^{\prime}(\bar{\tau})~=~&2\pi i\left(\frac{1}{24\bar{\tau}^2}+1\right),\cr S^{\prime \prime}(\bar{\tau})~=~&-\frac{\pi i}{6\bar{\tau}^3}.\end{align}

Note that

$$\varphi(-\lambda\bar{\tau}^{-1}) ~\to ~1 \qquad \text{for}\qquad\lambda~\to~\infty.$$

There are 2 stationary points:

\begin{align} \bar{\tau}_{\pm}~=~& \frac{\pm i}{2\sqrt{6}} \cr \Updownarrow&\cr \tau_{\pm}~=~& \frac{\pm i}{2\sqrt{6}\lambda} \cr \Updownarrow&\cr \tilde\tau_{\pm}~=~& \pm i 2\sqrt{6}\lambda. \cr \end{align}

We calculate

\begin{align} S(\bar{\tau}_{\pm})~=~&\mp\pi \sqrt{\frac{2}{3}}, \cr S^{\prime \prime}(\bar{\tau}_{\pm})~=~&\pm 8\sqrt{6}\pi.\end{align}

Clearly only the stationary point $$\bar{\tau}_+$$ in the upper halfplane contributes. The method of steepest descent yields

\begin{align} P(n) ~\sim~&\frac{1}{\lambda\sqrt{2\pi S^{\prime \prime}(\bar{\tau}_+)}} \frac{e^{-\lambda S(\bar{\tau}_+)}}{\sqrt{-i\bar{\tau}_+}} \cr ~=~&\frac{e^{\lambda \pi \sqrt{\frac{2}{3}}}}{\lambda2\pi\sqrt{2}}\qquad \text{for}\qquad\lambda~\to~\infty, \end{align}

and therefore

$$p(n)~=~P^{\prime}(n) ~\sim~\frac{e^{\lambda \pi \sqrt{\frac{2}{3}}}}{\lambda^2 4\sqrt{3}} \qquad \text{for}\qquad\lambda~\to~\infty.$$

For more details, see Ref. 1.

References:

1. E.M. Stein & R. Shakarchi, Complex Analysis, 2003; Appendix A.4, Theorem 4.1, p. 334.
• Thank you for this. I have looked into the proof by Stein and Shakrachi and am able to follow. However, I am confused about one step towards the end of the proof just before we apply the method of steepest descent. Stein obtains $$-\mu^{\frac{3}{2}}\int_{0}^{\pi}e^{2s\sin{(\theta)}}e^{\frac{3i\theta}{2}}\sqrt{i} \, d\theta = \mu^{\frac{3}{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{2s\cos{(\theta)}}\left(\cos{\left(\frac{3\theta}{2}\right)}+i\sin{\left(\frac{3\theta}{2}\right)}\right) \, d\theta.$$ How do we go from one integral to the other is what is confusing me at the moment. Commented Feb 20, 2021 at 15:55