Convert the Laplace transform of the Bessel function to a Fourier transform

Question

I want to calculate the Fourier transform of the function $f(t)$, defined as $f(t)=0$ if $t<0$ and $f(t)=J_{n}(t)$ if $t\ge0$, in which $J_{n}(t)$ is the Bessel function of the first kind. That is, I want to calculate $$F(\omega)=\int_0^\infty J_{n}(t)e^{-i\omega t}dt.$$ As far as I know, I can do it by calculating its Laplace transform and then setting $s=i\omega$. In Laplace transform of the Bessel function of the first kind, it is shown how to calculate it by solving the integral $$F(s)=\int_0^\infty J_n(bt)e^{-st}dt=\frac{1}{i\pi}\int_{\mid z \mid=1}\frac{z^n}{bz^2+2sz-b}dz$$ that has simple poles at $z=-\frac{s}{b}\pm\frac{\sqrt{s^2+b^2}}{b}$. As the OP says, only $z=-\frac{s}{b}+\frac{\sqrt{s^2+b^2}}{b}$ is inside the unit circle.

The thing is: if I set $s=i\omega$, the simple poles are $z=-\frac{i\omega}{b}\pm\frac{\sqrt{b^2-\omega^2}}{b}$, and now both poles are inside the unit circle. That is, if I take the Laplace transform and set $s=i\omega$, I get $$F_{1}(\omega)=\frac{(\sqrt{a^2+b^2}-a)^n}{b^{n}\sqrt{a^2+b^2}}\Big|_{s=i\omega}=\frac{\left(\sqrt{1-\left(\frac{\omega}{b}\right)^2}-i\frac{\omega}{b}\right)^n}{b\sqrt{1-\left(\frac{\omega}{b}\right)^2}}$$ However, if I solve directly the previous integral with $s=i\omega$, I get $$F_{2}(\omega)=\frac{\left(\sqrt{1-\left(\frac{\omega}{b}\right)^2}-i\frac{\omega}{b}\right)^n}{b\sqrt{1-\left(\frac{\omega}{b}\right)^2}}-\frac{\left(-\sqrt{1-\left(\frac{\omega}{b}\right)^2}-i\frac{\omega}{b}\right)^n}{b\sqrt{1-\left(\frac{\omega}{b}\right)^2}}$$ Curiously, the best numerical results I get when I use $F_{1}(\omega)$... So what am I getting wrong? Why should I drop the pole $z=-\frac{i\omega}{b}-\frac{\sqrt{b^2-\omega^2}}{b}$?

Additional info: I use the function $f(t)$ as a filter, that is, I use it to calculate $h(t)=f(t)\ast g(t)$ for a given signal $g(t)$. Although $F_{2}(\omega)$ seems to provide the correct Fourier transform when compared to the FFT of $f(t)$, the convolution is (a lot) inaccurate. However, the convolution is accurate when I use $F_{1}(\omega)$.

Answer 1

In that evaluation, I wrote \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, \mathrm dx &= \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\pi}^{\pi} e^{i(n \theta -bx \sin \theta)} e^{-ax} \, \mathrm d \theta \, \mathrm dx \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \int_{0}^{\infty} e^{i n \theta} e^{-(a+ib \sin \theta)x} \, \mathrm dx \, \mathrm d \theta \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta}}{a + ib \sin \theta} \, \mathrm d \theta. \end{align}

Assuming that $b$ is a positive value, this is only true if the real part of $a$ is greater than zero.

If $a$ is purely imaginary, then the integral $$\int_{0}^{\infty} e^{-(a+ib \sin \theta)x} \, \mathrm dx $$ does not converge to $\frac{1}{a+ib \sin \theta}$.

However, you can indeed replace $a$ with $i \omega, \ \omega \in \mathbb{R},$ because $$\int_{0}^{\infty} J_{n}(bx)e^{-i\omega x} e^{-px} \, \mathrm dx $$ converges uniformly for $p \in [0, \infty)$ as long as $b \ne |\omega|$. See my answer here. Also see this question.