In equation (9) of this paper, 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 $s\to0$?

  • 1
    $\begingroup$ I found a copy of "The H-Function: Theory and Applications" on the net and after a very short glance on the text after equation (2.8), that you cited, there were several restrictions on the parameters. One of them I interpret in a way that the formula for the transform only applies if $\alpha >2 n +1$, which is not the case. Did I miss something ? $\endgroup$ Jul 3, 2019 at 13:34

1 Answer 1


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.

  • $\begingroup$ Thanks for the succinct derivation. Can the series expansion technique in the question be used to instead evaluate the small-$s$ asymptotic behaviour of $$\int_0^\infty\frac{\cos(kx_0)}{s+Dk^\alpha}dk ~\text{?}$$ It works when $x_0=0$, since only the first (i.e., $n=0$) term in the series becomes nonzero, but this is no longer the case when $x_0\neq0$. $\endgroup$
    – user121642
    Jul 2, 2019 at 14:39
  • $\begingroup$ I think that also may work, but I think just using dominated convergence gives the result faster. $\endgroup$ Jul 2, 2019 at 17:03

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