I'm interested in the integral $$I=\int_{-\pi/2}^{\pi/2} W(\sec(\varphi)) d\varphi$$where $W$ is the Lambert W function, defined such that $W(x)e^{W(x)}=x$. The numerical value for $I$, as yielded by wolfram alpha, is $$2.824240767667472\ldots$$
I would love either a closed form representation of this value, or a proof that no such representation exists. I know that the indefinite integral has no solution in terms of standard functions, but I was hoping that the definite integral named just might.
An equivalent integral that may be a little easier (?) is $$I=2\int_{W(1)}^{\infty}\frac{1+x}{\sqrt{(xe^x)^2-1}}$$ which was arrived at by the substitution of $xe^x=\sec(\varphi)$, and converges extremely rapidly in the area in question.
And that's all I really have. I'm not quite sure how to attack this problem, and would love your insights. If nothing else, I hope you find it interesting to think about for a few minutes!
$\DeclareMathOperator\W W$This question has not been answered for years, so here is a series representation. Integrating the inverse of $\W(\sec(x))$, we get:
$$\int_{-\frac\pi2}^\frac\pi2\W(\sec(\varphi))d\varphi=\pi\W(1)+2\int_{\W(1)}^\infty\csc^{-1}(xe^x)dx$$
$\csc^{-1}(z)$ has a series representation which converges for $|z|>1$ giving $|xe^x|>1\iff x>\W(1)$ for $z=xe^x$:
$$\int_{\W(1)}^\infty\csc^{-1}(xe^x)dx=\sum_{n=1}^\infty\left(\frac12\right)_n\frac1{(2n+1)n!}\int_{\W(1)}^\infty(x e^x)^{-2n-1}dx$$
with the Pochhammer symbol and the incomplete gamma function:
$$\boxed{\int_{-\frac\pi2}^\frac\pi2\W(\sec(\varphi))d\varphi=\pi\W(1)+2\sum_{n=0}^\infty\left(\frac12\right)_n\frac{(2n+1)^{2n-1}}{n!}\Gamma(-2n,(2n+1)\W(1))}$$
Shown here:
which does not appear to have a closed form. Expanding $\Gamma(a,x)$ may reveal more forms.