Due to Rayleigh's formulas we have:
$$\sqrt{\frac{\pi}{2}}\frac{J_{k+1/2}(t)}{t^{k+1/2}}=(-1)^k\left(\frac{1}{t}\frac{d}{dt}\right)^k \frac{\sin t}{t}\tag{1}$$
and since:
$$\frac{\sin t}{t}=\sum_{m=0}^{+\infty}\frac{(-1)^m\,t^{2m}}{(2m+1)!}$$
we have:
$$\left(\frac{1}{t}\frac{d}{dt}\right)\frac{\sin t}{t}=\sum_{m=1}^{+\infty}\frac{(-1)^m(2m)t^{2m-2}}{(2m+1)!}=(-1)\sum_{m=0}^{+\infty}\frac{(-1)^m (2m+2)t^{2m}}{(2m+3)!},$$
$$\left(\frac{1}{t}\frac{d}{dt}\right)^k\frac{\sin t}{t}=(-1)^k\sum_{m=0}^{+\infty}\frac{(-1)^m (2m+2k)\cdot\ldots\cdot(2m+2)t^{2m}}{(2m+2k+1)!}$$
so:
$$\begin{eqnarray*}(2k+1)!!\cdot\sqrt{\frac{\pi}{2}}\frac{J_{k+1/2}(t)}{t^{k+1/2}}&=&\sum_{m=0}^{+\infty}\frac{(-1)^m (2m+2k)!!(2k+1)!!}{(2m+2k+1)!(2m)!!}\,t^{2m}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m \binom{m+k}{m}}{\binom{2m+2k+1}{2m}}\cdot\frac{t^{2m}}{(2m)!}.\end{eqnarray*}\tag{2}$$
Now a really good approximation for the LHS of $(2)$ is simply given by:
$$(2k+1)!!\cdot\sqrt{\frac{\pi}{2}}\frac{J_{k+1/2}(t)}{t^{k+1/2}}\approx \exp\left(-\frac{t^2}{4k+6}\right).\tag{3}$$
$\hspace2in$
$\hspace2in\qquad$ Approximation for $k=3$
Hence the starting integral can be approximated by:
$$\int_{0}^{+\infty}t\,\exp\left(-\frac{qt^2}{4k+6}\right)\,dt=\frac{2k+3}{q}.$$