Closed form of an integral involving Lambert function I'm trying to compute the following integral explicitly.
$$I=\int_{0}^{+\infty} dx \left(1+\frac{1}{x}\right) \frac{\sqrt{x}}{e^{-1}+xe^x}$$
The best I managed to do is to do a change of variable $x=W(y)$, where W is the Lambert function. The integral is then given by 
$$I=\int_{0}^{+\infty} \frac{dy}{y}  \frac{\sqrt{W(y)}}{e^{-1}+y}$$
Maybe there could be a way to deform the contour of integration in the complex plane and use a residue formula with the new contour as we could have a pole at $y=-e^{-1}$?
I guess we could do the transformation $y=e^x$ and obtain
$$I=\int_{-\infty}^{+\infty} dx  \frac{\sqrt{W(e^x)}}{e^{-1}+e^x}$$
The poles would be located at $x=-1\pm i \pi$ but I do not know which contour to choose...
I computed numerically the integral on mathematica which gives $I\simeq 3.9965$
 A: $$I = \int\limits_{0}^{+\infty} \frac{x+1}{e^{-1}+xe^x}\frac{dx}{\sqrt{x}} = 2e\int\limits_{0}^{+\infty} \frac{y^2+1}{1+y^2e^{y^2+1}}dy = e\int\limits_{-\infty}^{+\infty} \frac{y^2+1}{1+y^2e^{y^2+1}}dy .$$
Roots of the denominator to be found from the equality
$$y^2e^{y^2+1} = -1,$$
which has the solutions
$$y_n = \sqrt{W_n(-1/e)},\quad n \in\mathcal Z,$$
where $\ W_n\ $ is $n$-th branch of complex Lambert function.
Note that
$$\mathrm{Res}\left(\dfrac{1+y^2}{1+y^2e^{y^2+1}},\,y_n \right) = \lim_{y\to y_n}\dfrac{1+y^2}{2ye^{y^2+1}(1+y^2)} = \frac12\lim_{y\to y_n}\dfrac{y_n}{y_n^2e^{y_n^2+1}} = -\dfrac{y_n}2.\qquad(1)$$
Let us consider 
$$e^{-y^2}+ey^2=0$$
If $\ y=u+iv\ $ then $\ y^2 = u^2 - v^2 + 2iuv,$
$$e^{v^2-u^2-2iuv}= e(v^2-u^2-2iuv),$$
$$\begin{cases}
e^{v^2-u^2}\cos(2uv)= e(v^2-u^2)\\
e^{v^2-u^2}\sin(2uv)= 2euv,
\end{cases}$$
Using the main trig identity and the universal trig substitution, one can obtain
$$\begin{cases}
e^{v^2-u^2}= e(v^2+u^2)\\
\tan(uv)= \dfrac{2euv}{e^{v^2-u^2}+e(v^2-u^2)}
\end{cases}\rightarrow
\begin{cases}
e^{v^2-u^2}= e(v^2+u^2)\\
\tan(uv)= \dfrac uv
\end{cases}\rightarrow
\begin{cases}
v^2 = u^2 + \log(v^2+u^2) +1\\
uv = \pi n + \arctan\dfrac uv.
\end{cases}$$ 
This research show that for $n\to\infty$ must be
$$\mathcal Re\, y_n\approx \mathcal Im\,y_n\approx \sqrt{\pi\left(n+\frac14\right)}+O(n^{-1/2}\log\, n)$$
That means that the infinity sum of residues $(1)$ diverges.
