This summation is as far as I got with $L=\displaystyle\lim\limits_{a\to\infty}\frac{1}{a}\int\limits_{-\infty}^{a}\{e^x\}dx$:
$$L=\lim\limits_{a\to\infty}\frac{\sum\limits_{n=1}^{a}\left(1-n\ln\left(\frac{n+1}{n}\right)\right)}{\ln a}$$
I don't know how to progress from here.
I am assuming that by $\{x\}$ you mean the fractional part of $x$. First observe that the limit does exist since $$\int_{-\infty}^{a} \{e^x\} dx = \int_{-\infty}^{0}e^x dx + \int_0^{a} \{e^x\} dx \le 1 + a $$ and therefore $\frac{1}{a}\int_{-\infty}^{a} \{e^x\} dx \le 1+\frac{1}{a}$ for all $a>0$. Thus, $L = \lim_{n \to \infty} \frac{1}{\log(n+1)}\int_{-\infty}^{\log(n+1)} \{e^x\}dx$ and the integral can be simplified as $$ \begin{align*}\int_{-\infty}^{\log(n+1)} \{e^x\}dx &= 1+\sum_{k=1}^{n} \int_{\log(k)}^{\log(k+1)} (e^x - k) dx = 1+\sum_{k=1}^{n}\left. (e^x - kx) \right|_{\log(k)}^{\log(k+1)} \\ &= 1+\sum_{k=1}^{n} (1-k\log(1+1/k)) = 1+\sum_{k=1}^{n} \frac{1}{2k}+ \sum_{k=1}^{n} O\left(\frac{1}{k^2}\right)\end{align*} $$ We have, $$\frac{1}{\log(n+1)}\int_{-\infty}^{\log(n+1)} \{e^x\}dx = \frac{1}{2}\frac{\log(n)}{\log{(n+1)}} +O\left(\frac{1}{\log(n+1)}\right) \to \frac{1}{2} $$