I am trying to compare asymptotic runtime bounds of a few algorithms presented in this research paper, A quasi-polynomial algorithm for discretelogarithm in finite fields of small characteristic.  The functions are, $L(\alpha) = \exp(O(\log(n)^{a}(\log\log(n))^{1-\alpha}))$, which I'd like to plot for $\alpha = 1/3, 1/4+O(1)$, and $1/4 + O(n)$,and $O(n^{\log n})$ Where n is the bit-size of the input. ... Such a complexity is smaller than any $L(\epsilon)$ for any $\epsilon > 0$ I would like to plot each of these functions together to compare their growth.  The problem is that the second function grows much faster than the first, which implies I am misinterpreting something.So what is the correct way to interpret and compare these functions?
How to interpret these asymptotic runtime bounds for discrete logarithm algorithms?
asymptotics
The confusion is that in the expression for $L(\alpha)$, $n$ is size of the field, whereas in $n^{O(\log n)}$, $n$ is the size of the input. The correct comparison is between $L(\alpha) = \exp O((\log n)^\alpha (\log\log n)^{1-\alpha})$ and $(\log n)^{O(\log\log n)} = \exp O((\log\log n)^2)$, which is indeed much smaller.Note that your suggestion to plot $L(\alpha)$ for $\alpha = 1/4 + O(1)$ and $\alpha = 1/4 + O(n)$ doesn't make much sense; you should think of $\alpha$ as a constant. It could make sense to consider $\alpha = 1/4 + o(1)$ (or $\alpha = 1/4 + \epsilon$, where $\epsilon$ is understood to be small), though it's not clear how exactly you'd plot it. Indeed, there is no reason to plot, and a plot wouldn't make too much sense without knowing the hidden big O constant. In practice you just do experiments.