I am working on the following function for a fixed $0<\nu<1$. $$ f(t) = E_\nu (-t^\nu) = \sum \limits_{n=0}^\infty (-1)^n\frac{t^{n \nu}}{\Gamma(n\nu +1)} $$ where $E_\nu(\cdot)$ is a Mittag-Leffler function.
I want to show that
$f(t) = E_\nu (-t^\nu)$ is decreasing for $t>0$.
I have tried the usual idea: Showing the derivative is negative. However the derivative function which is $$ f'(t) = \frac{d}{dt}E_\nu (-t^\nu) = \sum \limits_{n=1}^\infty (-1)^n\frac{t^{n \nu-1}}{\Gamma(n\nu)} $$ also hard to deal with. I have tried separating the cases $0<t<1$ and $t >1$. I believe for large $t$ the derivative function should be "much more negative". I couldn't prove that either.
Any help will be appreciated.
Using the integral representation for $0<\nu<1$ (e.g. see this paper https://arxiv.org/pdf/1305.0161 , equation 2.14) $$ f(t)=E_\nu (-t^\nu) = \sum \limits_{n=0}^\infty (-1)^n\frac{t^{n \nu}}{\Gamma(n\nu +1)}=\frac{1}{\pi}\int_0^\infty e^{-xt}\frac{x^{\nu-1} \sin (\pi \nu)}{x^{2 \nu}+2 x^\nu \cos (\pi \nu)+1}dx $$ we immdediately see that $f(t)$ is decreasing for $t>0$.