(i) the formula in (2) is another consequence of the Sparre Andersen transformation, see e.g. Theorem 4.4 here (ii) "kind of rescale" leaves room for interpretation. I just meant the fact that exchanging the "scale parameter" $\frac{\delta^2}{\sigma^2}$ transforms one distribution into the other.

A proof of $\frac{1-\log(2)}{2-\log(\tfrac{36}{5})}\geq 8$ : the denominator is positive, thus we may equivalently show that $$15\leq 8\log(\tfrac{36}{5})-\log(2)= 23\log(2)+8\log(\tfrac{9}{10})\;\;.$$ $\log(\tfrac{9}{10})>-\tfrac{1}{9}$ (since $-\log(1-x)<\tfrac{x}{1-x}$ for $x\in(0,1)$), thus it suffices to show that $23\log(2) -\tfrac{8}{9}\geq 15$, i.e. that $\log(2)\geq \tfrac{143}{207}$. Taking the first two terms in $\log(2)=2\,\mathrm{artanh}(\tfrac{1}{3})=2\sum_{k=0}^\infty \frac{1}{(2k+1)\,3^{2k+1}}$ gives $\log(2)>\tfrac{56}{81}$, and $\frac{56}{81}> \frac{143}{207}$.

A simple description of that distribution is given in mathoverflow.net/questions/222705/…

Let $I_n$ denote the $n$-variable integral above. The representation \begin{align*} I_{n+1}=\frac{1}{n!} \int_0^\infty (F(z))^n\,\frac{e^{-z}}{z}\,dz\;\;, \end{align*} where $F(z)=\int_0^z \frac{1-e^{-y}}{y}\,dy$, may also be of interest. (It can be obtained by writing $\frac{1}{t_k+\ldots t_n}=\int_0^\infty e^{-(t_k+\ldots + t_n)x_k}\,dx_k$ for $k=1,..,n$, carrying out the $t_k$ integrals, and a symmetry argument).

I just have confirmed your calculation of that coefficient, and corrected my post.

@Johannes Trost: no offence, and thanks for confirming the expansions

Using probabilistic reasoning I get \begin{align*} C(x)\,e^{-x^2}&= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{1619583}{425984}x^{-8}+{O}(x^{-10})\\ L(x)\,x^2\,e^{-x^2}&= 1+\frac{15}{8}x^{-2} + \frac{665}{128}x^{-4}+\frac{19845}{1024}x^{-6}+\frac{3747582‌​3 }{425984}x^{-8}+{O}(x^{-10}) \end{align*} These differ from your first findings but seem to be close to your later findings (I haven't checked the numerics).

If the $a_i$ are probabilities you can use the inequality of Mallows, see jstor.org/stable/2334886

@i707107 Thanks, that's a useful addition. I suspect that the result for the uniform case is much older, but have been unable to spot it.

Yes, it is with regard to $n$.

The limiting distributions of the maximal and minimal loads are described in chapter 2, §6 of the book "Random allocations" by Kolchin, Sevast'yanov and Chistyakov