Asymptotic integration of $\int_0^\infty\frac{x^{-\frac{1}{2}+a}J_{-\frac{1}{2}+a}(x\alpha)}{e^x-1}{\rm d}x$ when $\alpha \gg 1$ What is the asymptotic integration of
$$\int_0^\infty\frac{x^{-\frac{1}{2}+a}J_{-\frac{1}{2}+a}(x\alpha)}{e^x-1}{\rm d}x$$
when $\alpha\rightarrow\infty$. How to compute that using standard identities? Please give a general expression and then consider a special case where $a=\frac{5}{2}$.
 A: We assume that $a>\frac{1}{2}$ is fixed. Note that
$$
\frac{{x^{a - 1/2} }}{{e^x  - 1}} \sim \sum\limits_{n = 0}^\infty  {\frac{{B_n }}{{n!}}x^{n + a - 3/2} } 
$$
as $x\to 0+$, where $B_n$ denotes the Bernoulli numbers. Then, by Theorem 2 in https://doi.org/10.1137/0507061, we find
\begin{align*}
\int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{e^x  - 1}}J_{a - 1/2} (\alpha x)dx}  & \sim \frac{1}{{2}}\sum\limits_{n = 0}^\infty  {\frac{{B_n }}{{n!}}\frac{\Gamma\! \left( {\frac{n}{2} + a - \frac{1}{2}} \right)}{\Gamma\! \left( {1-\frac{n}{2}} \right)}\!\left( {\frac{2}{\alpha }} \right)^{n + a - 1/2} } 
\\ &
 = \frac{1}{2}\Gamma \!\left( {a - \frac{1}{2}} \right)\!\left( {\frac{2}{\alpha }} \right)^{a - 1/2}  - \frac{{\Gamma (a)}}{{4\sqrt \pi  }}\left( {\frac{2}{\alpha }} \right)^{a + 1/2}  
\end{align*}
as $\alpha \to +\infty$. Note that the Bernoulli numbers of odd index $n\geq 3$ or the reciprocal gamma function eliminate all terms in the asymptotic expansion except the first two. The absolute error of this two-term approximation decays to zero faster in $\alpha$ than any negative power of $\alpha$. Thus, this asymptotics is rather accurate for large $\alpha$.
In the special case $a=\frac{5}{2}$, the aproximation is
$$
\int_0^{ + \infty } {\frac{{x^2 }}{{e^x  - 1}}J_2 (\alpha x)dx}  \sim \frac{2}{{\alpha ^2 }} - \frac{3}{{2\alpha ^3 }}.
$$
The right-hand side apporaches $0$ from above for large positive values of $\alpha$.
A different type of expansion may be obtained as follows. We can expand the denominator of the integrand and integrate term-by-term using http://dlmf.nist.gov/10.22.E49, to deduce
\begin{align*}
\int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{e^x  - 1}}J_{a - 1/2} (\alpha x)dx} & = \int_0^{ + \infty } {x^{a - 1/2} \left( {\sum\limits_{n = 1}^\infty  {e^{ - nx} } } \right)J_{a - 1/2} (\alpha x)dx} 
\\ &
 = \sum\limits_{n = 1}^\infty  {\int_0^{ + \infty } {x^{a - 1/2} e^{ - nx} J_{a - 1/2} (\alpha x)dx} } 
\\ &
 = \left( {\frac{\alpha }{2}} \right)^{a - 1/2} \Gamma (2a)\sum\limits_{n = 1}^\infty  {\frac{1}{{n^{2a} }}{\bf F}\!\left( {a,a + \frac{1}{2};a + \frac{1}{2}; - \left( {\frac{\alpha }{n}} \right)^2 } \right)} 
\\ & = \left( {\frac{\alpha }{2}} \right)^{a - 1/2} \frac{{\Gamma (2a)}}{{\Gamma \!\left( {a + \frac{1}{2}} \right)}}\sum\limits_{n = 1}^\infty  {\frac{1}{{(n^2  + \alpha ^2 )^a }}} 
\\ &
 = (2\alpha )^{a - 1/2} \frac{{\Gamma (a)}}{{\sqrt \pi  }}\sum\limits_{n = 1}^\infty  {\frac{1}{{(n^2  + \alpha ^2 )^a }}} 
\end{align*}
for $a>\frac{1}{2}$ and $\alpha>0$. Here $\mathbf F$ stands for the regularised hypergeometric function. The last series is a generalised Mathieu series. Thus applying Theorem 1 in https://arxiv.org/abs/1601.07751v1, we obtain
\begin{align*}
\int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{e^x  - 1}}J_{a - 1/2} (\alpha x)dx}  = & \;\frac{1}{2}\Gamma\! \left( {a - \frac{1}{2}} \right)\!\left( {\frac{2}{\alpha }} \right)^{a - 1/2}  - \frac{{\Gamma (a)}}{{4\sqrt \pi  }}\left( {\frac{2}{\alpha }} \right)^{a + 1/2}  \\ & +(2\pi )^{a - 1/2} \frac{{e^{ - 2\pi \alpha } }}{{\sqrt \alpha  }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{\alpha }} \right)} \right)
\end{align*}
as $\alpha \to +\infty$, with $a>\frac{1}{2}$ being fixed. This shows the exponential accuracy of our original two-term asymptotics.
