Let $M(z,\nu):= \frac{1}{\pi}\sum_{n=1}^{\infty} \frac{(-z)^{n-1}}{(n-1)!}\Gamma(\nu n)\sin(\nu n\pi)=\frac{1}{2\pi i}\int_{\text{H}_a}\exp(\sigma -z\sigma^{\nu})/\sigma^{1-\nu} d\sigma$, $0<\nu<1$ where $H_a$ is the Hankel contour.

I'm reading a paper which uses, as a black-box, the following asymptotic estimate for $r\in\mathbb{R}$:

$M(r;\nu)\sim \frac{1}{\sqrt{2\pi(1-\nu)}}r^{(\nu-1/2)/(1-\nu)}\exp(-\frac{1-\nu}{\nu}r^{1/(1-\nu)}) , r\rightarrow +\infty$.

I am looking for hints or proofs of the above asymptotic behaviour: any ideas on how one can derive it ? Is this a saddle-point integral approximation applied to the Hankel contour integral ?

Let $M(z;\nu):= \frac{1}{\pi}\sum_{n=1}^{\infty} \frac{(-z)^{n-1}}{(n-1)!}\Gamma(\nu n)\sin(\nu n\pi)=\frac{1}{2\pi i}\int_{\text{H}_a}\exp(\sigma -z\sigma^{\nu})/\sigma^{1-\nu} d\sigma$, $0<\nu<1$ where $H_a$ is the Hankel contour.

I'm reading a paper which uses, as a black-box, the following asymptotic estimate for $r\in\mathbb{R}$:

$M(r;\nu)\sim \frac{1}{\sqrt{2\pi(1-\nu)}}r^{(\nu-1/2)/(1-\nu)}\exp(-\frac{1-\nu}{\nu}r^{1/(1-\nu)}) , r\rightarrow +\infty$.

I am looking for hints or proofs of the above asymptotic behaviour: any ideas on how one can derive it ? Is this a saddle-point integral approximation applied to the Hankel contour integral ?

Let $M(z,\nu):= \frac{1}{\pi}\sum_{n=1}^{\infty} \frac{(-z)^{n-1}}{(n-1)!}\Gamma(\nu n)\sin(\nu n\pi)=\frac{1}{2\pi i}\int_{\text{H}_a}\exp(\sigma -z\sigma^{\nu})/\sigma^{1-\nu} d\sigma$, $0<\nu<1$ where $H_a$ is the Hankel contour.

I'm reading a paper which uses, as a black-box, the following asymptotic estimate for $r\in\mathbb{R}$:

$M(z;\nu)\sim \frac{1}{\sqrt{2\pi(1-\nu)}}r^{(\nu-1/2)/(1-\nu)}\exp(-\frac{1-\nu}{\nu}r^{1/(1-\nu)}) , r\rightarrow +\infty$.

I am looking for hints or proofs of the above asymptotic behaviour: any ideas on how one can derive it ? Is this a saddle-point integral approximation applied to the Hankel contour integral ?

Let $M(z,\nu):= \frac{1}{\pi}\sum_{n=1}^{\infty} \frac{(-z)^{n-1}}{(n-1)!}\Gamma(\nu n)\sin(\nu n\pi)=\frac{1}{2\pi i}\int_{\text{H}_a}\exp(\sigma -z\sigma^{\nu})/\sigma^{1-\nu} d\sigma$, $0<\nu<1$ where $H_a$ is the Hankel contour.

I'm reading a paper which uses, as a black-box, the following asymptotic estimate for $r\in\mathbb{R}$:

$M(r;\nu)\sim \frac{1}{\sqrt{2\pi(1-\nu)}}r^{(\nu-1/2)/(1-\nu)}\exp(-\frac{1-\nu}{\nu}r^{1/(1-\nu)}) , r\rightarrow +\infty$.

I am looking for hints or proofs of the above asymptotic behaviour: any ideas on how one can derive it ? Is this a saddle-point integral approximation applied to the Hankel contour integral ?