<s>[INST] Integration of a Bessel function times a sine function times a polynomial I have recently faced the following expression (while trying to compute the Fourier transform of a RKKY-like potential):
\begin{align}
\int_0^{\infty} \text{d} r\;  J_0 (k r) \frac{\sin \left(\alpha \sqrt{1+r^2} \right) }{(1+r^2)^2} \; r = -\frac{1}{\alpha} \int_0^{\infty} \text{d} r\;  J_0 (k r) \frac{1}{(1+r^2)^{3/2}} \frac{\text{d}}{\text d r} \left[\cos \left( \alpha\sqrt{1+r^2} \right) \right]
\end{align}
Although I can solve it numerically, I would like to find a closed-form expression, if there is one.
I found some related questions here, such as,
\begin{align}
\int_0^{\infty} \text{d}r\; J_0 \left( \alpha\sqrt{x^2 +z^2} \right) \, \cos \left(\beta x \right)  = \frac{\cos \left(z\sqrt{\alpha^2 -\beta^2} \right)}{\sqrt{\alpha^2 -\beta^2}}
\end{align}
but did not really helped me.
 [/INST] According to this reference, Table 8.2 , entry (7):
$$\int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}J_0(kr)r \; dr = e^{-k}$$
and according to Table 8.2, entry (41):
$$\int_0^\infty \dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr = \dfrac{\cos\left[\left(\alpha^2-k^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k^2\right)^\frac{1}{2}}\Pi\left(\dfrac{k}{\alpha}-\dfrac{1}{2}\right)$$
where
$$\Pi(t) = \begin{cases} 1 & -\frac{1}{2} < t< \frac{1}{2}\\ 0 & \mathrm{otherwise} \end{cases}$$
Using the convolution property of the Hankel Transform, if the convolution exists
$$\begin{align*} \int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}\dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr &= \dfrac{1}{2\pi}e^{-k}**\dfrac{\cos\left[\left(\alpha^2-k^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k^2\right)^\frac{1}{2}}\Pi\left(\dfrac{k}{\alpha}-\dfrac{1}{2}\right)\\
\\
&= \dfrac{1}{2\pi}\int_0^{2\pi}\int_0^{k\cos\theta+\sqrt{k^2\cos^2\theta-(k^2-\alpha^2)}} e^{-k'}\dfrac{\cos\left[\left(\alpha^2-\left[k^2+k'^2-2kk'\cos\theta\right]\right)^\frac{1}{2}\right]}{\left(\alpha^2-\left[k^2+k'^2-2kk'\cos\theta\right]\right)^\frac{1}{2}}k' \; dk' \; d\theta\\
\\
&\mathrm{or}\\
\\
&= \dfrac{1}{2\pi}\int_0^{2\pi}\int_0^{\alpha^-} e^{-\sqrt{k^2+k'^2-2kk'\cos\theta}}\dfrac{\cos\left[\left(\alpha^2-k'^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k'^2\right)^\frac{1}{2}}k' \; dk' \; d\theta\\
\end{align*}$$
Neither of which look possible to easily evaluate symbolically.  Although maybe one of them takes less effort to evaluate numerically than your original integral.
</s>