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Claim 1: The function $$\frac{e^{1-ex}}{h_{2n}(e^{-ex})}+\frac{e^{x/e-1}}{h_{2n}(e^{x/e})}dx$$ is monotonically increasing in $n$ for $0<x<1$ and for $n>1$.

Claim 1: The function $$\frac{e^{1-ex}}{h_{2n}(e^{-ex})}+\frac{e^{x/e-1}}{h_{2n}(e^{x/e})}dx$$ is monotonically increasing in $n$ for $0<x<1$ and for $n>3$.

There are a few ways to clean up this second integral, such as $$\int_{e^{-e}}^{e^{1/e}}\frac{-\log x}{W(-\log x)}dx = \int_{\frac{1}{e}}^{e}\frac{1}{x}x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)dx$$ $$=\int_{-1}^{1}e^{-2u}e^{e^{-u}}(1-u)du.$$

There are a few ways to clean up this second integral, such as $$\int_{e^{-e}}^{e^{1/e}}\frac{-\log x}{W(-\log x)}dx = \int_{\frac{1}{e}}^{e}\frac{1}{x}x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)dx$$ $$=\int_{-1}^{1}e^{-2u}e^{e^{-u}u}(1-u)du.$$

Proposition 1: We have that $$\lim_{n\rightarrow \infty} \int_0^\infty \frac{1}{h_{2n}(x)} dx = 1.9187310\dots $$.

In particular, due to the monotonicity on the different ranges, Lemma 2 implies uniform convergence, and hence $$\lim_{n\rightarrow \infty}\int_0^{e^{1/e}} \frac{1}{h_{2n}(x)}dx=\int_{0}^{e^{1/e}}L(x)dx.$$

Proposition 2: We have that $$\lim_{n\rightarrow \infty}\int_0^{\infty} \frac{1}{h_{2n}(x)}dx=\int_{0}^{e^{1/e}}L(x)dx.$$ $$ = \int_{0}^{e^{-e}}L(x)dx + \int_{e^{-e}}^{e^{1/e}}\frac{-\log x}{W(-\log x)}dx.$$

The integral $\int_{0}^{e^{1/e}}L(x)dx$ does not seem to have as nice a form. To evaluate it, I wrote a program in Python that computes a Riemann sum with an error tolerance of $10^{-8}$.

Proposition 1: We have that $$\lim_{n\rightarrow \infty} \int_0^\infty \frac{1}{h_{2n}(x)} dx = C_h=1.91873106\dots $$ where $C_h$ may be expressed exactly as $$C_h=e^{1−e}+\int_{-1/e}^e\frac{e^{-u}u}{W(u)}du-\int_1^{e}e^{uW_{-1}[-\log u/u]} du$$ where $W$ and $W_{-1}$ are branches of the W-Lambert Function.

Thank you Fred Hucht for pointing out in the comments how to explicitely state the integral involving $W_{-1}$.

In particular, due to the monotonicity on the different ranges, Lemma 2 implies uniform convergence, and hence $$\lim_{n\rightarrow \infty}\int_0^{e^{1/e}} \frac{1}{h_{2n}(x)}dx=\int_{0}^{e^{1/e}}\frac{1}{L(x)}dx.$$

Proposition 2: We have that $$\lim_{n\rightarrow \infty}\int_0^{\infty} \frac{1}{h_{2n}(x)}dx=\int_{0}^{e^{1/e}}\frac{1}{L(x)}dx.$$ $$ = \int_{0}^{e^{-e}}\frac{1}{L(x)}dx + \int_{e^{-e}}^{e^{1/e}}\frac{-\log x}{W(-\log x)}dx.$$

Fred Hucht pointed out in the comments that the integral $\int_{0}^{e^{1/e}}\frac{1}{L(x)}dx$ can be expressed in terms of $W_{-1}(x)$, a different branch of the W-Lambert function, in the following way:

$$\int_{0}^{e^{-e}}\frac{1}{L(x)}dx = e^{1−e}-\int_1^{e}e^{uW_{-1}[-\log u/u]} du.$$ This implies the exact expression: $$\lim_{n\rightarrow \infty}\int_0^{\infty} \frac{1}{h_{2n}(x)}dx = e^{1−e}+\int_{-1/e}^e\frac{e^{-u}u}{W(u)}du-\int_1^{e}e^{uW_{-1}[-\log u/u]} du.$$