Monotonicity of a function involving Mittag-Leffler Function

Question

The Mittag-Leffler Function is defined as \begin{equation*} E_{a,b} : x \mapsto \sum_{k=0}^\infty \frac{x^k}{\Gamma(ak+b)} \end{equation*} Let's look at the function \begin{equation*} f_a : x \in \mathbb{R}_+ \mapsto E_{a,a}(t^a) \end{equation*} for $0<a<1$. I want to show, that \begin{equation*} x >0 \mapsto x^{a-1} e^{-x} f_a(x) \end{equation*} is non-increasing (plotting the function works). Differentiating this function and rearranging shows, that it is suffice to show, that \begin{equation*} (x+1)f_a(x) \geq ax^a \sum_{k=1}^\infty \frac{kx^{ak}}{\Gamma(ak+a)} \end{equation*} (which looks true by plotting). How can i show this?

Answer 1

Your function (we assume $0<a<1$ and $x>0$ throughout) $$g_a(x)=x^{a-1}e^{-x}f_a(x)=x^{-1}e^{-x}\sum_{n=1}^\infty\frac{x^{na}}{\Gamma(na)}$$ has the following integral representation: $$g_a(x)=\frac1a+\frac{\sin a\pi}{\pi}\int_0^\infty\frac{t^a e^{-x(1+t)}\,dt}{1-2t^a\cos a\pi+t^{2a}}$$ (which makes the non-increase obvious).

This representation may be obtained using the formula $$\frac1{\Gamma(s)}=\frac1{2\pi i}\int_\lambda z^{-s}e^z\,dz\qquad(s\in\mathbb{C})$$ where the contour $\lambda$ encircles the negative real axis, with its immediate corollary $$\frac{x^{s-1}}{\Gamma(s)}=\frac1{2\pi i}\int_\lambda z^{-s}e^{xz}\,dz.$$ Suppose that $\lambda_1$ encircles $|z|=1$ together with the negative real axis. Then $$g_a(x)=\frac{e^{-x}}{2\pi i}\sum_{n=1}^\infty\int_{\lambda_1}z^{-na}e^{xz}\,dz=\frac1{2\pi i}\int_{\lambda_1}\frac{e^{x(z-1)}}{z^a-1}\,dz.$$ Now use $\int_{\lambda_1}(\ldots)\,dz=\int_{\lambda_2}(\ldots)\,dz+2\pi i\operatorname*{res}\limits_{z=1}(\ldots)$, where $\lambda_2$ encircles the negative real axis closely (staying in $\Re z<1$). In our case, the residue equals $1/a$. And, in the limit of "closely", we get $$g_a(x)=\frac1a+\frac1{2\pi i}\int_0^\infty\left(-\frac{e^{x(-t-1)}}{t^a e^{-a\pi i}-1}+\frac{e^{x(-t-1)}}{t^a e^{a\pi i}-1}\right)d(-t),$$ which simplifies to the representation stated above.