I have a following integral $$\mathcal{I}=\int_0^\infty \mathop{dx}x^{-m}\frac{j_2(ax)}{(ax)^2},$$ where $j_2(x)$ is the spherical bessel function, $a>0$ is a real parameter and $m>1$. You do not need to worry about the convergence of the integral; let us for a moment assume that it converges.
I did the following change of variable: $$y=ax,$$ so that $dy=a\mathop{dx}$. In that situation, my integral becomes \begin{equation}\mathcal{I}=a^{m-1}\int_0^\infty \mathop{dx}y^m\frac{j_2(y)}{(y)^2},\end{equation} so that $$\mathcal{I}\sim a^{m-1}.$$ Now, here's the real problem: the above is definitely not valid when $a=0$. But, when $a\rightarrow 0$, but still $\neq 0$, the way I rewrote $\mathcal{I}$ should still be okay, which should give me $\mathcal{I}\rightarrow 0.$ But, this is not right. When $a\rightarrow 0$, $$\frac{j_2(ax)}{(ax)^2}\rightarrow \frac{1}{15},$$ which should give me $$\mathcal{I}=\frac{1}{15}\int_0^\infty \mathop{dx}x^{-m}.$$ I don't understand why the two approaches are giving me different results. What would be the correct way to rewrite the integral so that my $a$ comes out of the integral? Any ideas would be appreciated.
We may remove the parameter $a$ by a change of variable and get: $$ I(a) = \int_{0}^{+\infty}\frac{j_2(ax)}{(ax)^2}x^{-m}\,dx = a^{m-1}\int_{0}^{+\infty}\frac{j_2(x)}{x^{m+2}}\,dx = a^{m-1}\sqrt{\frac{\pi}{2}}\int_{0}^{+\infty}\frac{J_{5/2}(x)}{x^{m+3/2}}\,dx,$$ then the last integral can be computed through the Laplace transform, by exploiting: $$ \mathcal{L}(J_m(x))=\frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^m},\qquad \mathcal{L}^{-1}\left(\frac{1}{x^{m+3/2}}\right)=\frac{s^{m+1/2}}{\Gamma\left(\frac{3}{2}+m\right)}$$ from which: $$ I(a) = \frac{a^{m-1}}{\Gamma\left(\frac{3}{2}+m\right)}\sqrt{\frac{\pi}{2}}\int_{0}^{+\infty}\frac{s^{m+1/2}\,ds}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^{5/2}} $$ and by replacing $s$ with $\sinh u$ we have: