Inspired by this question on Mathematica.SE
In the linked question, the OP defines the function, $F(v)$, as
$$F(v) = -v + \frac{1}{1+\frac{v}{e^v-1}}-\frac{1}{1+\frac{ve^v}{e^v-1}}$$
and uses this function to define a normalisation constant:
\begin{align} I & = \int^{+\infty}_{-\infty}{e^{\int^x_0{F(v)}\:dv}}\:dx \\ & \approx 2.87581846048 \end{align}
I've been trying to find a closed form for this constant, but have gotten stuck on evaluating the integral in the exponential, let alone the outer integral. Is it possible to evaluate either of these two integrals using only real methods, or is complex analysis required?
This is not an answer.
Using what @Roman answered in the linked post, increasing the working precision, the constant seems to be $$\sim 2.8758184604794044915$$ which is not recognized by inverse symbolic calculators.
Thanks to a friend of mine who enjoys this kind of problems, the approximation $$\frac{640+726 \pi -431 \pi ^2}{142-975 \pi +249 \pi ^2}\sim 2.8758184604794044898$$
Edit
I do not think that we could obtain a closed form. However, if you look at the plot of $F(v)$, it is quite close to a straight line. Expanding as Taylor series around $v=0$, we have $$F(v)=-\frac{3 v}{4}-\frac{v^3}{192}+O\left(v^5\right)$$ which is an underestimate. $$g(x)=\int_0^x F(v)\,dv=-\frac{3 x^2}{8}-\frac{x^4}{768}$$ $$\int_{-\infty}^{\infty}e^{g(x)}\,dx=6 \sqrt{2}\, e^{27/2} \,K_{\frac{1}{4}}\left(\frac{27}{2}\right)\approx 2.87506$$
A little bit better using Padé approximant $$F(v)\sim -\frac{\frac{3 v}{4}+\frac{v^3}{120}}{1+\frac{v^2}{240}}$$ which is equivalent to a $O\left(v^7\right)$ expansion. $$g(x)=\int_0^x F(v)\,dv=-x^2+150 \log \left(1+\frac{x^2}{240}\right)$$ $$\int_{-\infty}^{\infty}e^{g(x)}\,dx=\sqrt{\pi b }\, U\left(\frac{1}{2},a+\frac{3}{2},b\right)\quad \text{with}\quad a=150\,,\,b=240$$ Evaluated, this gives $2.87538$.