I have the following function, where $E_{a,b}(.)$ is the Mittag-Leffler two paremeter function:
$$f(x)=x^{a-1}E_{a,a}(-x^a)$$ for $0<a\leq 1$.
Well, a people says that $f(x)\sim x^{-1-a}$ for large $x$.
I could not understand this. For instance, if $a=1$, $f(x)=-x^{2a-1}=-x\neq x^{-1-a}=x^{-2}$, what do you think?
I'd like to know the correct assumption and how can I prove this.
Many thanks!
PS.: How is the pronunciation of 'Mittag-Leffler'?
Best Answer
The Wikipedia article states the asymptotics of $E_{a,b}(z)$ for $z\to-\infty$, but the argument given there, for when and why it holds, isn't clear for me (regarding why, it seems to rely on the behaviour of $E_{a,b-ka}(z)$ as $z\to-\infty$, which looks like a circular dependence). A rigorous treatment may be based on the contour integral representation, which is proved using the Hankel's $\frac{1}{\Gamma(s)}=\frac{1}{2\pi\mathrm{i}}\int_\lambda z^{-s}e^z\,dz$ (where the contour $\lambda$ encircles the negative real $z$-axis, and the principal value of $z^{-s}$ is taken), valid for all complex values of $s$, and Watson's lemma (or its complex extension if needed).
Specifically, for real $x>0$ and $0<a<1$, we have (again with the p.v. of $z^a$) $$f_a(x):=x^{a-1}E_{a,a}(-x^a)=\frac{1}{2\pi\mathrm{i}}\int_\lambda\frac{e^{xz}\,dz}{1+z^a};\tag{*}$$ making $\lambda$ encircle closely the negative real axis, in the limit of "closely", we get $$f_a(x)=\frac{1}{2\pi\mathrm{i}}\int_0^\infty\left(\frac{1}{1+e^{-\mathrm{i}\pi a}t^a}-\frac{1}{1+e^{\mathrm{i}\pi a}t^a}\right)e^{-xt}\,dt$$ (substituting $z=e^{\pm\mathrm{i}\pi}t$ on the upper/lower edge of $\lambda$ with $+/-$ respectively). For $t<1$, we have $$\frac{1}{2\pi\mathrm{i}}\left(\frac{1}{1+e^{-\mathrm{i}\pi a}t^a}-\frac{1}{1+e^{\mathrm{i}\pi a}t^a}\right)=\frac1\pi\sum_{n=1}^{\infty}(-1)^{n-1}t^{na}\sin n\pi a,$$ and a variant of Watson's lemma (for a power series in $t^a$) gives $$f_a(x)\asymp\frac1\pi\sum_{n=1}^{(\infty)}(-1)^{n-1}\sin n\pi a\,\frac{\Gamma(na+1)}{x^{na+1}}=\bbox[5px,border:2px solid]{\sum_{n=1}^{(\infty)}\frac{(-1)^n}{\Gamma(-na)}\frac{1}{x^{na+1}}.}$$
This gives what you've been told for $0<a<1$. At $a=1$, the entire series vanish, and in fact many things change, starting with a note that $(\textrm{*})$ still holds if $\lambda$ encircles the disc $|z|\leqslant 1$, not only the negative real axis (in this case, $a>1$ is also admissible); the asymptotic analysis similar to the above is still possible, but it must be done more carefully. The same pertains to complex values of $x$ (however, still with $\Re x>0$).