Closed form of an integral involving Lambert function

Solution 1:

$$I = \int\limits_{0}^{+\infty} \frac{x+1}{e^{-1}+xe^x}\frac{dx}{\sqrt{x}} = 2e\int\limits_{0}^{+\infty} \frac{y^2+1}{1+y^2e^{y^2+1}}dy = e\int\limits_{-\infty}^{+\infty} \frac{y^2+1}{1+y^2e^{y^2+1}}dy .$$ 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.