I'm trying to understand how to get from one equation to another in a certain paper I am studying (DOI:10.1080/00018738100101467, eqs. 4.34 and 4.35). The equations are pretty self contained, so I'm not gonna waste time giving the context. The first equation is
$W(\beta)=\frac{2\beta}{\pi}\intop_{0}^{\infty}d\eta\,\eta\sin(\beta\eta)\exp\left(-\eta^{3/2}c\right)$
and the paper says that you can easily obtain an asymptotic series for this by expanding the exponential and integrating term-by-term. They get
$W(\beta)=\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-c)^{n-1}}{n!}\frac{\Gamma(\frac{3}{2}n+2)}{\beta^{3n/2+1}}\sin(\frac{3}{4}\pi n)$
I'm not sure how they got this. When you expand the exponential, you get integrals of the form
$\int d\eta\,\eta^{1+3n/2}\sin(\beta\eta)$
which as far as I know, cannot be evaluated. From the series they got, it appears that this integral is proportional to
$\Gamma(\frac{3}{2}n+2) \sin(\frac{3}{4}\pi n)$
but I cannot find any identity for Gamma functions which says this! Does anyone know how they got their series result?
0. Summary There is a rigorous aproach, which yields a valid series expansion not in the question, and the authors' approach to the series expansion in the question, which cannot be recommended.
1. A convergent series Expanding the sine, one gets $W(\beta)$ as the sum of series, namely, $$W(\beta)=\frac2\pi\sum_{n=1}^\infty(-1)^{n+1}\frac{\beta^{2n}w_n(c)}{(2n-1)!},\qquad w_n(c)=\int_{0}^{\infty}\eta^{2n}\exp\left(-\eta^{3/2}c\right)\,d\eta.$$ The change of variable $x=\eta^{3/2}c$ yields $\eta=c^{-2/3}x^{2/3}$ and $d\eta=\frac23c^{-2/3}x^{-1/3}dx$ hence $$w_n(c)=\int_{0}^{\infty}\frac23c^{-2/3}x^{-1/3}c^{-4n/3}x^{4n/3}\mathrm e^{-x}\,dx=\frac23c^{-(4n+2)/3}\Gamma((4n+2)/3),$$ which yields the series $$ W(\beta)=\frac4{3\pi}\sum_{n=1}^\infty(-1)^{n+1}\frac{\Gamma((4n+2)/3)}{\Gamma(2n)}\frac{\beta^{2n}}{c^{(4n+2)/3}}. $$ By the ratio test, this series converges absolutely for every $c\ne0$.
2. A divergent series By the ratio test, the series proposed in the question diverges for every $c\ne0$. It might be obtained as follows. First, the change of variable $\beta\eta=x^2$ yields $\beta d\eta=2xdx$ and $$W(\beta)=\frac{4}{\pi\beta}V\left(\frac{c}{\beta^{3/2}}\right),\qquad V(a)=\int_{0}^{\infty}x^3\sin(x^2)\exp\left(-ax^3\right)dx.$$ Next, expanding the exponential in $V(a)$ and pretending that one can exchange the summation and the integral (one cannot), one gets $$V(a)=\sum_{n=0}^\infty\frac{(-a)^n}{n!}v_{3n+3},\qquad v_k=\int_{0}^{\infty}x^k\sin(x^2)dx.$$ Finally, I guess the authors have a (nonmathematical) way to attribute a value to each of the diverging integrals $v_k$ and to sum them...