In equation (9) of [this paper](https://arxiv.org/pdf/cond-mat/0309449.pdf), it is claimed that the limiting behaviour
$$
\int\_0^\infty \frac{1-\cos(kx\_0)}{s+Dk^\alpha}dk
\sim
\frac{\Gamma(2-\alpha)\sin(\pi(2-\alpha)/2)x\_0^{\alpha-1}}{(\alpha-1)D} \tag{1}\label{papereq}
$$


holds as $s\to0$, when $1<\alpha<2$.
However, by using the series expansion of $\cos(kx\_0)$
$$
\cos(kx\_0) = \sum\_{n=0}^\infty\frac{(-1)^n(kx\_0)^{2n}}{(2n)!} ~,
$$
along with the identity (Appendix p152, *The H-function with Applications in Statistics and Other Disciplines*)
$$
\frac{z^\beta}{1+az^\alpha}=a^{-\beta/\alpha}H\_{1,1}^{1,1}\left[az^\alpha\Bigg|\begin{matrix}(\beta/\alpha,1)\\(\beta/\alpha,1)\end{matrix}\right] ~,
$$
we get
$$
\int\_0^\infty \frac{1-\cos(kx\_0)}{s+Dk^\alpha}dk\\
=
\sum\_{n=1}^\infty\frac{(-1)^{n+1}x\_0^{2n}(D/s)^{-2n/\alpha}}{(2n)!\alpha s}
\int\_0^\infty dk\_1k\_1^{1/\alpha-1}H\_{1,1}^{1,1}\left[\frac{D}{s}k\_1\Bigg|\begin{matrix}(2n/\alpha,1)\\(2n/\alpha,1)\end{matrix}\right] ~,
$$
and using the Mellin transform of the Fox $H$-function (equation (2.8), *The H-Function: Theory and Applications*) gives
$$
\int\_0^\infty \frac{1-\cos(kx\_0)}{s+Dk^\alpha}dk\\
=
\sum\_{n=1}^\infty\frac{(-1)^{n+1}x\_0^{2n}(D/s)^{-2n/\alpha}}{(2n)!\alpha s}
(D/s)^{-1/\alpha} \Gamma(2n/\alpha+1/\alpha)\Gamma(1-2n/\alpha-1/\alpha) ~.
$$
Taking the first term of the series gives us
$$
\int\_0^\infty \frac{1-\cos(kx\_0)}{s+Dk^\alpha}dk
\sim
\frac{x\_0^{2}(D/s)^{-3/\alpha}}{(2)!\alpha s}\Gamma(3/\alpha)\Gamma(1-3/\alpha) ~. \tag{2}
$$


However, this expression cannot be reconciled with $\eqref{papereq}$, especially considering that the powers of $D$ and $x\_0$ have differing dependences on $\alpha$ between the two equations.
This leads one to believe that this process of obtaining asymptotic approximations doesn't work with these expressions.


What might be the correct way to obtain asymptotic approximations for this expression, as $s\to0$?



Asymptotic behaviour of function using Fox $H$-function representation
By dominated convergence, as $s\downarrow0$, 
\begin{align\*}
\int\_0^\infty \frac{1-\cos(kx\_0)}{s+Dk^\alpha}dk
&\to
\int\_0^\infty \frac{1-\cos(kx\_0)}{Dk^\alpha}dk \\
&=\frac{x\_0^{\alpha-1}}D\,\int\_0^\infty \frac{1-\cos u}{u^\alpha}du \\ 
&=-\frac{x\_0^{\alpha-1}}D\,\Gamma(1 - \alpha) \sin(\pi \alpha/2) \\
&=\frac{\Gamma(2-\alpha)\sin(\pi(2-\alpha)/2)x\_0^{\alpha-1}}{(\alpha-1)D}. 
\end{align\*}


Note: To evaluate $\int\_0^\infty \frac{1-\cos u}{u^\alpha}du$, one may, for instance, integrate by parts and then use formula 3.712.1 (page 419) or 17.43.4 (page 1131) in [Gradshteyn and Ryzhik](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwiIh8O4sJbjAhVEJ80KHfC2CS0QFjACegQICRAC&url=http%3A%2F%2Ffisica.ciens.ucv.ve%2F~svincenz%2FTISPISGIMR.pdf&usg=AOvVaw01xV-Pb5aQAUw6cJvOGtpR). 


