In a dual integral situation, the following integral has to be involved $$ \int_0^\infty \frac{\cos (qt) J_1 (qr)}{1+q^2} \, \mathrm{d} q \quad\quad (0<t<r) \, . $$
Visibly this integral is convergent. I was wondering whether an amenable analytical expression is possible? This will be useful for my further analysis.
Any help is highly appreciated.
Thanks.
R
On
Let $H_{1}^{(1)}(z)$ be the Hankel function of the first kind of order one defined as $$H_{1}^{(1)}(z) = J_{1}(x) + i Y_{1}(x). $$
Similar to the answer here, we can exploit the fact that $$H_{1}^{(1)}(xe^{i \pi}) = J_{1}(x) - i Y_{1}(x), \quad x>0,$$ to write
$$\int_{0}^{\infty} \frac{\cos(tq) J_{1}(rq)}{1+q^{2}} \, \mathrm dq = \frac{1}{2} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{\cos(tq) H_{1}^{(1)}(rq)}{1+q^{2}} \, \mathrm dq. $$
Let's integrate the function $$f(z) = \frac{\cos(tz) H_{1}^{(1)}(rz)}{1+z^{2}}, \quad 0 < t \le r, $$ around a contour that consists of the upper side of the branch cut on the negative real axis, a small semicircle about the origin ($C_{\epsilon}$), the positive real axis, and a large semicircle in the upper half-plane.
The integral along the large semicircle vanishes as the radius of the semicircle goes to infinity. This is due the fact that the magnitude of the $\cos(tz) H_{1}^{(1)}(rz)$ is bounded in the upper half-plane if $0<t \le r$.
Therefore, we have
$$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{\cos(tq)H_{1}^{(1)}(rq)}{1+q^{2}} \, \mathrm dq + \lim_{\epsilon \to 0} \int_{C_{\epsilon}}f(z) \, \mathrm dz = 2 \pi i \operatorname{Res}[f(z), i],$$ where
$$\begin{align} 2 \pi i \operatorname{Res}[f(z), i] &= 2 \pi i \left(\frac{\cosh(t) H_{1}^{(1)}(ir)}{2i} \right) \\ &\overset{(1)}{=} \pi \cosh(t) \lim_{\nu \to 1} i \csc(\pi \nu) \left(e^{-i \pi \nu} J_{\nu}(ir) - J_{-\nu}(ir) \right) \\ &\overset{(2)}{=} i \pi \cosh(t) \lim_{\nu \to 1} \csc(\pi \nu) \left(e^{-i \pi \nu /2} I_{\nu}(r) - e^{- i \nu \pi /2} I_{-\nu}(r) \right) \\ & = - \pi \cosh(t) \lim_{\nu \to 1} \csc(\pi \nu) \left(I_{-\nu}(r) - I_{\nu}(r) \right) \\ & \overset{(3)}{=} - 2 \cosh(t) K_{1}(r). \end{align} $$
But since $$\lim_{z \to 0} (z-0) \, \frac{\cosh(tz) H_{1}^{(1)}(rz)}{1+z^{2}} = i \lim_{z \to 0} z Y_{1}(rz) \overset{(4)}{=} \frac{2}{i \pi r},$$ it follows that $$\lim_{\epsilon \to 0} \int_{C_{\epsilon}}f(z) \, \mathrm dz = - i \pi \left( \frac{2}{i \pi r} \right) = -\frac{2}{r}. $$
(This is basically the fractional residue theorem except the singularity at the origin is a branch point and not a simple pole.)
Putting everything together, we get $$\int_{0}^{\infty} \frac{\cos(tq) J_{1}(rq)}{1+q^{2}} \, \mathrm dq = \frac{1}{2} \, \Re \left(\frac{2}{r} - 2 \cosh(t) K_{1}(r) \right) = \frac{1}{r} - \cosh(t) K_{1}(r), \quad 0 <t \le r . $$
$(1)$ https://dlmf.nist.gov/10.4.E7
$(2)$ https://dlmf.nist.gov/10.27.E6
$(3)$ https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_I%CE%B1,_K%CE%B1
$(4)$ https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms
A similar approach exploiting the fact that $$H_{2n}^{(1)}(xe^{i \pi}) = -J_{2n}(x) + iY_{2n}(x), \quad (x >0, \ n \in \mathbb{Z}_{\ge 0} ),$$ shows that $$\int_{0}^{\infty} \frac{\sin(tq) J_{0}(rq)}{1+q^{2}} \, \mathrm dq = \sinh(t) K_{0}(r), \quad 0 < t \le r, $$ and
$$\int_{0}^{\infty} \frac{\sin(tq) J_{2}(rq)}{1+q^{2}} \, \mathrm dq = \frac{2t}{r^{2}} - \sinh(t) K_{2}(r) , \quad 0 < t \le r. $$
Following @tired's idea and using two known integral identities, we can compute the integral. Fix $r > 0$ and consider $I$ defined by
$$I(t) = \int_{0}^{\infty} \frac{\cos (tq) J_1 (rq)}{1+q^2} \, dq. $$
On the interval $(0, r)$, it satisfies the following 2nd ODE
$$ I(t) - I''(t) = \int_{0}^{\infty} \cos(tq)J_1(rq) \, dq, \qquad I(0) = \int_{0}^{\infty} \frac{J_1 (rq)}{1+q^2} \, dq, \quad I'(0) = 0. $$
We have two extra unknown integrals, but they can be computed using DLMF 10.22.59 and DLMF 10.22.46: for $0 < t < r$,
$$ \int_{0}^{\infty} \cos(tq)J_1(rq) \, dq = \frac{1}{r} \quad \text{and} \quad \int_{0}^{\infty} \frac{J_1 (rq)}{1+q^2} \, dq = \frac{1}{r} - K_1(r) \tag{*}$$
Thus the problem boils down to solving
$$ I(t) - I''(t) = \frac{1}{r}, \qquad I(0) = \frac{1}{r} - K_1(r), \quad I'(0) = 0. $$
Now the general solution of this equation is of the form
$$ I(t) = \frac{1}{r} + A \cosh t + B \sinh t $$
and plugging the initial condition shows $A = -1$ and $B = 0$. Therefore
$$ \int_{0}^{\infty} \frac{\cos (tq) J_1 (rq)}{1+q^2} \, dq = \frac{1}{r} - K_1(r) \cosh t, \qquad 0 < t < r. $$
p.s. I would love to see a self-contained solution as I don't quite understand $\text{(*)}$.