closed form evaluation of $\int_{-1}^{1} (x;x)_{\infty}\,dx$ This was mostly provoked by the graph on http://reference.wolfram.com/language/ref/QPochhammer.html which suggests
that the integral
$$\int_{-1}^1 (x;x)_\infty\,dx\approx 1.28830088867\ldots$$
is convergent. Is it convergent to something tidy?
I've used $x$ instead of $q$ here because $q$ is usually associated with being in the unit disc on the complex plane.
Seeing the fractional part start with $0.288$ is somewhat suggestive, given that $(1/2,1/2)_{\infty}=0.2887880950866024\ldots$ (entry 25 on http://mathworld.wolfram.com/TreeSearching.html)
 A: Result
The result of the integral
$$i = \int_{-1}^1 (x;x)_\infty\,dx $$
where
$$(x;x) = \prod _{m=1}^{\infty } \left(1-x^m\right)\tag{1}$$
can be obtained in closed form as follows
$$i = s = s_1 + s_2$$
where
$$ s_1 =2 \sum _{k=-\infty }^{\infty } \frac{1}{24 k^2+2 k+1} = -\frac{1}{23} i \left(\sqrt{23} \pi  \cot \left(\frac{1}{24} \left(\pi -i \sqrt{23} \pi \right)\right)-\sqrt{23} \pi  \cot \left(\frac{1}{24} \left(\pi +i \sqrt{23} \pi \right)\right)\right)\tag{2}$$
which can be simplified to
$$s_{1a} =\frac{2 \pi  \sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{23} \left(\cosh \left(\frac{\sqrt{23} \pi }{12}\right)-\frac{1+\sqrt{3}}{2 \sqrt{2}}\right)}\tag{2a}$$
and
$$s_2=2 \sum _{k=-\infty }^{\infty } \frac{1}{-24 k^2+34 k-13} = -\frac{1}{23} i \left(\sqrt{23} \pi  \tan \left(\frac{1}{24} \left(5 \pi -i \sqrt{23} \pi \right)\right)-\sqrt{23} \pi  \tan \left(\frac{1}{24} \left(5 \pi +i \sqrt{23} \pi \right)\right)\right)\tag{3} $$
which can be simplified to
$$s_{2a} = -\frac{2 \pi  \sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{23} \left(\frac{\sqrt{3}-1}{2 \sqrt{2}}+\cosh \left(\frac{\sqrt{23} \pi }{12}\right)\right)}\tag{3a}$$
Numerically,
$$N(s_1) = 2.2680458411340673629 ...$$ 
$$N(s_2) = -0.97974495246014613277...$$
$$N(s) = 1.28830088867392123018... $$
Derivation
Using Euler's theorem of pentagonal numbers [1] we can write the product as a sum
$$\prod _{m=1}^{\infty } \left(1-x^m\right)=\sum _{n=-\infty }^{\infty } (-1)^n x^{\frac{1}{2} n (3 n+1)}\tag{4}$$
The right hand side can be integrated term by term giving the result as one infinite sum. This in turn can be decomposed into two sums with indices $n = 4k$ and $n = 4k-3$.
Related material can be found in [2] and [3].
References
[1] http://mathworld.wolfram.com/PartitionFunctionP.html
Further reading
[2] https://arxiv.org/pdf/math/0505373.pdf, L.Euler (1775), On the remarkable properties of the pentagonal numbers
[3] MSE: How to prove Euler's pentagonal theorem? Some hints will help
