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The paper "Infinite integrals involving Bessel functions by contour integration" by Qiong-Gui Lin gives a method to solve integrals of the form $\intop_{0}^{\infty}x^{v}f(x)J_{v}(qx)\, dx$. One of the restrictions on $f(x)$ is that it is "practically a rational function, though theoretically may be beyond that". This is stated in the introduction of the method, but the actual theorem states

Theorem 2.1 Assume that $f(z)$ is analytic on the complex plane except for several singularities; $f(−z) = −f(z)$; on the positive real axis, it has only poles such that they are at most simple poles of $f(z)J_v(qz)$; $z^{2v+1}f(z) → 0$ uniformly when $z → 0$ and $z^{ν−1/2}f(z) → 0$ uniformly when $z→∞$; then, we have....

after which it gives the result (too complicated to post here).

I want to apply this theorem to the function $f(x)=\frac{x^{3}e^{-ax^{2}}}{c^{2}+x^{4}}$ (with $v=0$). This is not a rational function, but it does seem to satisfy all of the criteria given in the actual theorem. I quickly looked over the proof of the theorem, and I couldn't see any time where $f$ had to be rational.

The reason for my concern is that the resulting function does not display some of the properties I observed with asymptotics and numerics (such as a high $c$ limit of $I\sim c^{-2}$). Can I apply this theorem to my problem?

(I've asked about this integral before at Difficult infinite integral involving a Gaussian, Bessel function and complex singularities, let me know if it's bad to make another post)

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Ah, I guess the problem is that $f$ blows up on the imaginary axis, so $z^{v-1/2}f(z)$ does not go to 0 uniformly as $z$ goes to infinity.

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