Is there a way to compute $\int_0^\infty \frac{\cos (qt) J_1 (qr)}{1+q^2} \, \mathrm{d} q$ provided that $0In 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
 A: 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 $\text{(*)}$.
A: Using the clever approach from this answer, we have $$\int_{0}^{\infty} \frac{\cos(tq) J_{1}(rq)}{1+q^{2}} \, \mathrm dq = \frac{1}{2} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{\cos(tq) H_{1}^{(1)}(rq)}{1+q^{2}} \, \mathrm dq, $$ where here $H_{1}^{(1)}(x) = J_{1}(x) + i Y_{1}(x)$ is the Hankel function of the first kind of order $1$.
Let's integrate the function $$f(z) = \frac{\cos(tz) H_{1}^{(1)}(rz)}{1+z^{2}}, \quad 0 < t \le r, $$ around a large semicircular contour in the upper half of the complex plane that includes a small clockwise-oriented semicircle about the origin ($C_{\epsilon})$.
The integral along the big 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

Original answer:
The following is a way to show that $$I(0)= \int_0^\infty \frac{J_{1}(rq)}{1+q^2} \, \mathrm dq = \frac{1}{r} - K_{1}(r) \, , \quad r >0. $$
This integral appears in Sangchul Lee's answer.
Using the fact that $$ \int_{0}^{\infty} e^{-ax} \, \mathrm dx = \frac{1}{a}, \quad a >0, $$ along with the series representation of the Bessel function of the first kind, we have
$$ \begin{align} \int_0^\infty \frac{J_{1}(rq)}{1+q^2} \, \mathrm dq &= \int_0^\infty J_{1}(rq) \int_0^\infty e^{-(1+q^{2})t} \, \mathrm dt \, \mathrm dq \\ &= \int_0^\infty e^{-t} \int_0^\infty e^{-tq^{2}} J_{1}(rq)  \, \mathrm dq \, \mathrm dt \\ &=  \int_0^\infty e^{-t} \int_0^\infty e^{-tq^{2}}  \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{rq}{2} \right)^{2m+1} \, \mathrm dq \, \mathrm dt \\ &=  \int_0^\infty e^{-t}    \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{r}{2} \right)^{2m+1}\int_0^\infty e^{-tq^{2}} q^{2m+1} \, \mathrm dq \, \mathrm dt \\ &=\int_0^\infty e^{-t}    \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{r}{2} \right)^{2m+1} \frac{1}{2t^{m+1}}\int_0^\infty e^{-u} u^{m} \, \mathrm du \, \mathrm dt \\ &= \frac{1}{r}\int_0^\infty e^{-t}   \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(m+1)!} \left(\frac{r^2}{4 t} \right)^{m+1} \, \mathrm d t \\ &=\frac{1}{r}\int_0^\infty e^{-t}\left(1- \exp \left(-\frac{r^2}{4 t} \right) \right) \, \mathrm dt \\ &= \frac{1}{r} - \frac{1}{r} \int_0^\infty \exp \left(-t- \frac{r^{2}}{4t} \right) \, \mathrm dt  \\ & \overset{(1)}{=} \frac{1}{r} - \frac{1}{r} \, 2 \sqrt{\frac{r^2}{4}} K_{1}\left(2 \sqrt{\frac{r^{2}}{4}} \right) \\ &= \frac{1}{r}- K_{1}(r).\end{align}$$
$(5)$ How to find the integral $\int_{0}^{\infty}\exp(- (ax+b/x))\,dx$?
