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In , the Euler function is given by

\phi(q)=\prod_{k=1}^\infty (1-q^k),\quad |q|<1.
Named after , it is a model example of a and provides the prototypical example of a relation between and .


Properties
The p(k) in the formal power series expansion for 1/\phi(q) gives the number of partitions of k. That is,
\frac{1}{\phi(q)}=\sum_{k=0}^\infty p(k) q^k
where p is the partition function.

The Euler identity, also known as the Pentagonal number theorem, is

\phi(q)=\sum_{n=-\infty}^\infty (-1)^n q^{(3n^2-n)/2}.

(3n^2-n)/2 is a pentagonal number.

The Euler function is related to the Dedekind eta function as

\phi (e^{2\pi i\tau})= e^{-\pi i\tau/12} \eta(\tau).

The Euler function may be expressed as a q-Pochhammer symbol:

\phi(q) = (q;q)_{\infty}.

The of the Euler function is the sum of the logarithms in the product expression, each of which may be expanded about q = 0, yielding

\ln(\phi(q)) = -\sum_{n=1}^\infty\frac{1}{n}\,\frac{q^n}{1-q^n},

which is a with coefficients -1/ n. The logarithm of the Euler function may therefore be expressed as

\ln(\phi(q)) = \sum_{n=1}^\infty b_n q^n

where b_n=-\sum_{d|n}\frac{1}{d}= -1/1, (see A000203)

On account of the identity \sigma(n) = \sum_{d|n} d = \sum_{d|n} \frac{n}{d} , where \sigma(n) is the , this may also be written as

\ln(\phi(q)) = -\sum_{n=1}^\infty \frac{\sigma(n)}{n}\ q^n.

Also if a,b\in\mathbb{R}^+ and ab=\pi ^2, thenBerndt, B. et al. "The Rogers–Ramanujan Continued Fraction"

a^{1/4}e^{-a/12}\phi (e^{-2a})=b^{1/4}e^{-b/12}\phi (e^{-2b}).


Special values
The next identities come from Ramanujan's Notebooks:
(1998). 9781461272212, Springer.
p. 326

\phi(e^{-\pi})=\frac{e^{\pi/24}\Gamma\left(\frac14\right)}{2^{7/8}\pi^{3/4}}

\phi(e^{-2\pi})=\frac{e^{\pi/12}\Gamma\left(\frac14\right)}{2\pi^{3/4}}

\phi(e^{-4\pi})=\frac{e^{\pi/6}\Gamma\left(\frac14\right)}{2^

\phi(e^{-8\pi})=\frac{e^{\pi/3}\Gamma\left(\frac14\right)}{2^{29/16}\pi^{3/4}}(\sqrt{2}-1)^{1/4}

Using the Pentagonal number theorem, exchanging sum and , and then invoking complex-analytic methods, one derives

\int_0^1\phi(q)\,\mathrm{d}q = \frac{8 \sqrt{\frac{3}{23}} \pi \sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi }{3}\right)-1}.

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