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 $a=\frac{5}{2}$.
We assume that $a>\frac{1}{2}$ is fixed. Note that $$ \frac{{x^{a - 1/2} }}{{\mathrm{e}^x - 1}} \sim \sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}x^{n + a - 3/2} } $$ as $x\to 0^+$, where $B_n$ denotes the Bernoulli numbers. Then, by Theorem $2$ in this paper, we find \begin{align*} \int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{\mathrm{e}^x - 1}}J_{a - 1/2} (\alpha x)\,\mathrm{d}x} & \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 $$ \int_0^{ + \infty } {\frac{{x^2 }}{{\mathrm{e}^x - 1}}J_2 (\alpha x)\,\mathrm{d}x} \sim \frac{2}{{\alpha ^2 }} - \frac{3}{{2\alpha ^3 }}. $$ 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 $(10.22.49)$, to deduce \begin{align*} \int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{\mathrm{e}^x - 1}}J_{a - 1/2} (\alpha x)\,\mathrm{d}x} & = \int_0^{ + \infty } {x^{a - 1/2} \left( {\sum\limits_{n = 1}^\infty {\mathrm{e}^{ - nx} } } \right)J_{a - 1/2} (\alpha x)\,\mathrm{d}x} \\ & = \sum\limits_{n = 1}^\infty {\int_0^{ + \infty } {x^{a - 1/2} \mathrm{e}^{ - nx} J_{a - 1/2} (\alpha x)\,\mathrm{d}x} } \\ & = \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 this paper, we obtain \begin{align*} \int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{\mathrm{e}^x - 1}}J_{a - 1/2} (\alpha x)\,\mathrm{d}x} = & \;\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{{\mathrm{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.