$\log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr)+ \frac{1}{2} \log t$

Question

Let $t$ be a real $>0$.

Let $P(x) = \prod_{n=1}^\infty \dfrac{1}{1-x^n}$

$\log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr) + \frac{1}{2} \log t$

How to prove this without complex numbers ?

Answer 1

Your $P(x)$ is the multiplicative inverse of the Euler function and a generating function for the partition function : $$\tag{1} P(x)\ =\ \frac 1{\phi(x)}\ =\ \sum_{n=0}^\infty p(n)x^n$$

(in the following I'll replace '$x$' by the more common '$q$' letter)

It may too be written in function of the Dedekind eta function
(from $\ \phi(q)=q^{-\frac 1{24}}\eta(\tau)\ $ with $\ q=e^{2\pi i\tau}\ $) : $$\tag{2} P\bigl(e^{2\pi i\tau}\bigr)=\frac {e^{-\pi i\tau/12}}{\eta(\tau)}$$ (applied to $\tau:=it\,$ and $\,\tau:=i/t$)

after that you may use the eta functional equation : $$\tag{3} \ \eta(-1/\tau)=\ \sqrt{-i\tau}\ \,\eta(\tau)$$

or... use Weil or Siegel's proofs provided here by Paul Garrett...

---

ADDITION
At this point it should be clear that your equality is equivalent to the eta functional equation (in the real case) so don't hope a trivial proof.

Let's provide some details of Siegel's proof (or rather of a variant from Berndt and Venkatachaliengar's paper "On the transformation formula for the Dedekind eta-function" and Ayoub's book "An Introduction to Analytic Theory of Numbers").

We start with $$P(q) = \prod_{n=1}^\infty \dfrac{1}{1-q^n}$$ that we will rewrite as :

$$ \begin{align} \log P(q)&=-\sum_{n=1}^\infty \log (1-q^n)\\ &=\sum_{n=1}^\infty\ \sum_{k=1}^\infty\frac {q^{nk}}{k}\\ &=\sum_{k=1}^\infty \frac 1k\,\sum_{n=1}^\infty q^{nk}\\ &=\sum_{k=1}^\infty \frac 1{k(q^{-k}-1)}\\ \end{align} $$

For $\ q:=e^{-2\pi t}\ $ and $\ q:=e^{-2\pi/t}\ $ respectively this becomes :

$$\log P\left(e^{-2\pi t}\right)=\sum_{k=1}^\infty \frac 1{k(e^{2\pi kt}-1)}$$ $$\log P\left(e^{-2\pi/t}\right)=\sum_{k=1}^\infty \frac 1{k(e^{2\pi k/t}-1)}$$

getting :

$$F(t):=\log P\left(e^{-2\pi t}\right)-\log P\left(e^{-2\pi/t}\right)=\sum_{k=1}^\infty \frac 1k\left(\frac 1{e^{2\pi kt}-1}-\frac 1{e^{2\pi k/t}-1}\right)$$ Now $\ \displaystyle\frac 12 \left(\coth(x)-1\right)=\frac 1{e^{2x}-1}$ so that : $$F(t)=\sum_{k=1}^\infty \frac {\coth(\pi k t)-\coth(\pi k/t)}{2k}$$

Here point we could use residus (the quick end of Siegel's proof provided in Garrett's paper) but we will continue without complex with Berndt and Venkatachaliengar's variant :

We have $\ \displaystyle\pi\coth(\pi z)= \frac 1z+\sum_{n=1}^\infty \frac {2z}{n^2+z^2}$ (this and the $\pi\cot(\pi z)$ variant appears often here, note that the proof in the link only uses Fourier series) so let's set $z:=kt$ and $z:=k/t$ to get :

$$F(t)=\sum_{k=1}^\infty \frac 1{2\pi k} \left(\frac 1{kt}-\frac tk+\sum_{n=1}^\infty \frac {2kt}{n^2+k^2 t^2}-\frac {2k/t}{n^2+k^2/t^2}\right)$$ of course $\ \displaystyle\sum_{n=1}^\infty \frac 1{k^2}=\frac{\pi^2}6$ so that we get the first part of the wished $\frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr)+ \frac 12 \log(t)$ : $$F(t)=\frac{\pi}{12}\left(\frac 1t -t\right)+\frac 1{\pi}\sum_{k=1}^\infty \sum_{n=1}^\infty \left(\frac 1{n^2/t+k^2 t}-\frac 1{n^2 t+k^2/t}\right)$$

There is a temptation to use symmetry and conclude that the double series terms cancel but the more internal series has to be summed first. The proof of $$\frac 1{\pi}\sum_{k=1}^\infty \sum_{n=1}^\infty \left(\frac 1{n^2/t+k^2 t}-\frac 1{n^2 t+k^2/t}\right)=\frac 12 \log(t)$$ is a little long and may be consulted in Berndt's paper (starting with $(5)$ there and supposing that $\,a:=\pi t\,$ and $\,b:=\pi/t\,$).

Providing a shorter proof of this last part would of course be interesting!

---

Poisson summation formula: Let's obtain the eta functional equation from the Poisson summation formula (from Andrews, Askey and Roy's "Special functions" $(10.12.13-16)$).

Applied to the function $f(x)=e^{-s\pi x^2}$ the Poisson summation formula gives : $$\tag{1}\sum_{n=-\infty}^\infty e^{-s\pi(n+x)^2}=s^{-1/2} \sum_{n=-\infty}^\infty e^{-\pi n^2/s}e^{2\pi i n x}$$

Setting $\,q:=e^{\pi i\tau}\,$ and $\,p:=e^{-\pi i/\tau}\,$ this becomes, using the definition of the $\theta_3$ theta function : $$\tag{2}\theta_3(z,q)=\sum_{n=-\infty}^\infty q^{n^2}e^{2n i z}$$

$$\tag{3}\theta_3\left(\frac{\pi z}{\tau},p\right)=\sqrt{\frac {\tau}i}\ e^{\pi i(\pi z)^2/\tau}\ \theta_3(\pi z,q)$$ But using the q-Pochhammer notation we have : $$\tag{4}\theta_3(z,q)=(q^2;q^2)_\infty (-qe^{2iz};q^2)_\infty (-qe^{-2iz};q^2)_\infty$$ This infinite product gives us : $$\tag{5}\lim_{\lambda\to 1}\frac{\theta_3((\tau+\lambda)\pi/2,q)}{1+e^{-\lambda\pi i}}=\prod_{n=1}^\infty (1-q^n)^3$$ and $$\tag{6}\lim_{\lambda\to 1}\frac{\theta_3((\tau+\lambda)\pi/2\tau,p)}{1-p^{1-\lambda}}=\prod_{n=1}^\infty (1-p^n)^3$$ From $(3),(5),(6)$ we get : $$\eta(-1/\tau)^3=\frac{\tau}i\sqrt{\frac{\tau}i}\eta(\tau)^3$$ or $$\eta(-1/\tau)=c\sqrt{\frac{\tau}i}\eta(\tau)$$ with $c^3=1$. Set $\tau=i$ to get $c=1$.

I'll let you verify all that (it's late!).

Answer 2

Hints:

$$ \ln \prod_{n=1}^\infty f(n) = \sum_{n=1}^{\infty} \ln(f(n)), $$

$$ \ln\left( \frac{a}{b}\right) = \ln(a)-\ln(b) $$