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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 ?

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up vote 1 down vote accepted

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...

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!).

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hmmm... reading the 'without complex numbers' part again it seems that at least one 'undressing' will be necessary... ;-) – Raymond Manzoni Nov 23 '12 at 22:07
If you mean by undressing splitting complex in real and imag parts then I think that is necc yes. I do not want to use complex numbers or the eta functional equation. something 'simple' or 'classical' if possible. – mick Nov 23 '12 at 22:10
Can we use Poisson summation to prove the eta functional equation ? – mick Nov 27 '12 at 22:41
@mick: I updated my answer (not really straightforward either...). – Raymond Manzoni Nov 28 '12 at 1:03
Thanks @Raymond – mick Dec 29 '12 at 20:06


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

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

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I know that. That is trivial. Does not seem to help me. – mick Nov 23 '12 at 21:10

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