Substituting asymptotics into an integral?

Question

Let $J_{\nu}$ denote the Bessel function of the first kind. Then, for sufficiently large, positive $z$, we have

$$\displaystyle J_{\nu}(z) = O(z^{-1/2}).$$

Now suppose I want to integrate this Bessel function. Let $k \in \mathbb{R}^2$ be non-zero, and let $\rho$ be large but independent of both $k$ and $r$. Is the following valid?

$$\displaystyle \int_{0}^{|k|}|J_{1}(\rho r)| \ \mathrm{d}r \leqslant C\rho^{-1/2}\int_{0}^{|k|}r^{-1/2} \ \mathrm{d}r = C\rho^{-1/2}|k|^{1/2}.$$

Here, $C$ is some constant, and $\rho \to \infty$. This seems a bit naive, especially since the integral starts at $0$, which could mean my asymptotics aren't valid. Am I allowed to argue like this?

Note: I am aware that this integral can be done directly (the anti-derivative of the integrand is $-J_{0}(x)\text{sgn}(x)$). However, I'd just like to know if the idea works, so I can apply it to more complicated scenarios.

Answer 1

If $z\ll 1$ we have (by Taylor series) $$ J_{\nu}(z)\approx \frac{1}{\Gamma(\nu+1)}\left(\frac{z}{2}\right)^\nu \tag{1}$$ and if $z\gg \nu^2$ we have (by Laplace's method) $$ J_{\nu}(z)\approx \sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\nu \pi}{2}-\frac{\pi}{4}\right)\tag{2} $$ Obviously, if we need to approximate an integral in a right neighbourhood of the origin we may only use $(1)$, since $(2)$ is not granted to hold. We may even say $(2)$ cannot hold in a neighbourhood of the origin, since $J_\nu(z)$ does not have a singularity at $z=0$ (it is an entire function!).

For practical purposes, it is often convenient to recall that $$ \mathcal{L}(J_1(x)) = \frac{1}{\sqrt{1+s^2} \left(s+\sqrt{1+s^2}\right)} \tag{3} $$ and exploit the Cauchy-Schwarz inequality, too.