I need to demonstrate that the following function is nonnegative for $\alpha\in(1,2)$ and $\delta > 1$
$$ \sum_{m=0}^\infty {(-t^2/4)^m \over m!} \cdot { \Gamma^\prime \left( { \alpha +1 \over \alpha}+ m { 1 \over \alpha} \right) +\delta \Gamma \left( { 1\over \alpha} +m { 2\over \alpha}\right) \over \Gamma \left( { 1\over 2} +m\right) }, $$
where $\Gamma(\cdot)$ is a Gamma functions. One can recognize Fox-Wright function in the second term.
Here is the background. I would like to show that the Fourier transform of ${1 \over \alpha^{\delta} } \exp(-|x|^\alpha)$ is nonincreasing in $\alpha$. Taking derivative, $$ {\partial \over \partial \alpha } \left\{ {1 \over \alpha^{\delta} } \int_0^{\infty} \exp(-x^\alpha)\cos(xt) \, dx \right\}= -{1 \over \alpha^\delta} \int_0^\infty \exp(-x^\alpha)\{x^\alpha \log(x) + \delta/\alpha\}\cos(xt) \, dx $$
Using Taylor series expansion for $\cos(\cdot)$ leads to
\begin{align} &\int_0^\infty \exp(-x^\alpha)\{x^\alpha \log(x) + \delta/\alpha\}\cos(xt) \, dx \\ &= {1\over \alpha^2} \sum_{m=0}^{\infty} { (-t^2)^{m}\over (2m) !} \Gamma \left( { 2m+1\over \alpha}\right) \left\{\gamma\left( { 2m+1\over \alpha}\right) \left( { 2m+1\over \alpha}\right) + 1+\delta \right\} \\ &={1\over \alpha^2} \sum_{m=0}^{\infty} { (-t^2)^{m}\over (2m) !} \left[\Gamma^{\prime} \left( { 2m+1\over \alpha}+1\right) +\delta \Gamma \left( { 2m+1\over \alpha}\right) \right], \end{align}
where $\Gamma(\cdot)$ and $\gamma(\cdot)$ are gamma and digamma functions respectively.
There is nothing special about $\delta$, I found it numerically. The connection to Fox Wright function is
\begin{align} \int_0^\infty \exp(-x^\alpha)\cos(xt) \, dx &= {1\over \alpha} \sum_{m=0}^\infty \Gamma\{(1+2m)/\alpha\} { (-t^2)^{m}\over (2m) !} \\ &={\sqrt{\pi}\over \alpha} \sum_{m=0}^{\infty} { \Gamma\{(1+2m)/\alpha\}\over \Gamma(1/2+m)} { (-t^2/4)^ m \over m !}. \end{align}
Any hints/ references will be greatly appreciated as well.