In section 2.3.5 of Green, Schwarz, Witten's book on string theory (volume-1) pp. 116-118, the objective is to calculate an Asymptotic Formula for Level Densities $d_n$ for open bosonic string theory. $d_n$ is related to the classical partition function $f(w)$ via the generating function $$G(w)=\sum_{n=0}^\infty d_n w^n$$ and $$G(w)=f(w)^{-24}.$$ Two estimates are found out in the book in equations $(2.3.112)$ and $(2.3.117)$ respectively which we state here for convenience
- $$f(w) \sim \exp\left(-\frac{\pi^2}{6(1-w)}\right).\tag{2.3.112}$$
- $$f(w) \sim (1-w)^{-1/2}\exp\left(-\frac{\pi^2}{6(1-w)}\right).\tag{2.3.117}$$
Using this and the integral formula for Laurent series coefficients $$d_n=\frac{1}{2 \pi i}\oint dw \frac{G(w)}{w^{n+1}} \tag{2.3.118}$$ one can find out $d_n$ after making a saddle point approximation ($w \to 1$ as $n \to \infty$) as explained in the book. The final result in eqn. $(2.3.120)$ is
$$d_n \sim n^{-27/4} \exp(4 \pi \sqrt{n}).\tag{2.3.120}$$
I am able to find out the exponential piece which comes directly from the exponential piece in $f(w) \sim \exp\left(-\frac{\pi^2}{6(1-w)}\right)$, but not the extra $n^{-27/4}$ factor.
I thought that maybe then the more accurate approximation $(2.3.117)$ is to be used. But that will change the saddle points and what I am finding is a factor of $n^{-6}$. Given that the approximation $(2.3.117)$ is not even mentioned in a newer book ("Becker-Becker-Schwarz" pg. 52 subsection "The number of states") by one of the authors seems to imply that the first approximation $(2.3.112)$ is used to obtain the correct $n$ factor. But how? So in a nutshell my question is: Which approximation formula is being used in the texts and how does it give a factor of $n^{-27/4}$ in $(2.3.120)$?
ADDENDUM
This is how I got the exponential piece: The integrand in $(2.3.118)$ can be rewritten using $(2.3.112)$ as $$\frac{G(w)}{w^{n+1}}=\frac{1}{w^{n+1}}f(w)^{-24}\sim \frac{1}{w^{n+1}}\exp\left(\frac{4 \pi^2}{1-w}\right)=\exp \left(-(n+1) \ln(w) \right)\exp\left(\frac{4 \pi^2}{1-w}\right)$$ $$ \Rightarrow \frac{G(w)}{w^{n+1}}=\exp\left(\frac{4 \pi^2}{(1-w)}-(n+1) \ln(w)\ \right)$$ Now as $w \sim 1$, using Taylor expansion of $\ln(1-w)$ we can write $$-\ln(w)=-\ln(1-(1-w))\sim (1-w)$$ Therefore putting $-\ln(w)$ instead of $(1-w)$ in the first term of the argument of the exponential in the equation for $G(w)/w^{n+1}$, we get $$ \frac{G(w)}{w^{n+1}}=\exp\left(-\frac{4 \pi^2}{\ln(w)}-(n+1) \ln(w)\ \right) \tag{2.3.119}$$ From this the saddle point of the integrand is easily evaluated by differentiating w.r.t $w$ and setting the result to zero which gives $ \ln(w) \sim -2 \pi/\sqrt{n}$ as the saddle point, substituting which back to $(2.3.119)$ readily gives the exponential part of $(2.3.120)$. Doing a similar analysis using the approximation $(2.3.117)$ instead yields $n^{-6}$ instead of the expected $n^{-27/4}$.