How to study large $r \gg 1$ asymptotics of $$I(r):=\int_0^{\infty} \frac{1-e^{-q}}{1+q} J_0(rq) \ dq,$$ where $J_0$ is the zeroth order Bessel function of the first kind.
I did some numerics and it seems to indicate that $I(r) \sim 1/r^3$ for large $r$, but this is not obvious to me where it comes from.
To obtain an asymptotic expansion for large $r$, one might proceed as follows ($H_0$ is Struve function, $Y_0$ is Bessel function): $$I(r)=\int_0^{\infty} \frac{1-e^{-q}}{1+q} J_0(rq) \ dq=\int_0^{\infty} \frac{1}{1+q} J_0(rq)-\sum_{n=0}^\infty(-1)^n\int_0^{\infty} q^ne^{-q} J_0(rq)$$ $$\qquad=\tfrac{1}{2}\pi {H}_0(r)-\tfrac{1}{2}\pi Y_0(r)-\sum_{n=0}^\infty(-1)^n\Gamma (n+1) \, _2F_1\left(\frac{n+1}{2},\frac{n+2}{2};1;-r^2\right)$$ $$\qquad=\frac{3}{2 r^3}-\frac{123}{8 r^5}+{\cal O}(r^{-7}).$$
This is confirmed by a calculation using the representation derived here of the Bessel function integral as an integral of elementary functions, $$I(r)=\int_0^\infty dt\,e^{-t} \left(\frac{1}{\sqrt{ t^2+r^2}}-\frac{1}{\sqrt{( t+1)^2+r^2}}\right)$$ $$\qquad=\frac{3}{2 r^3}-\frac{123}{8 r^5}+\frac{6185}{16 r^7}-\frac{2424835}{128 r^9}+{\cal O}(r^{-11}).$$
