While conducting an analysis of an inverse Fourier transform stemming from a fluid mechanics problem related to flows in porous media, I encountered the following infinite integral:
$$
f(r) = \int_0^\infty \frac{k}{k^2+\alpha^2} \, J_0(k) J_0(kr) \, \mathrm{d}k \, ,
$$
where $r \ge 0$ and $\alpha > 0$. It's evident that the integrand behaves as $\mathcal{O}\left( k^{-2} \right)$ as $k \to \infty$, indicating the convergence of the integral.
My attempted approach involved utilizing the classical expression of the zeroth-order Bessel function:
$$
J_0(u) = \frac{1}{2\pi} \int_0^{2\pi} \exp \left(-iu \sin t \right) \, \mathrm{d} t
$$
I applied this expression to one as well as to both Bessel functions and attempted to evaluate them. Unfortunately, my efforts have not led to a successful result.
If anyone here can offer guidance or provide hints that might assist in evaluating this integral, I would greatly appreciate it.
E D I T
Based on numerical tests, it appears that for $\alpha = 1$, $f(r)$ exhibits a proportionality to the modified Bessel function of the second kind, $K_0(r)$. The approximate proportionality coefficient appears to be around $1.2660658\dots$.
For other values of $\alpha$, the trend is less discernible. It's possible that the behavior is a combination of multiple modified Bessel functions of the second kind, but this is a speculative hypothesis derived from intuition and not necessarily a reflection of reality.
Best Answer
UPDATE:
A generalization of this integral appears on page 429 of the textbook A Treatise on the Theory of Bessel Functions, namely
$$\small \int_{0}^{\infty} \frac{x}{x^{2}+\alpha^{2}} \, J_{\mu} (\beta x) \left(\cos \left(\frac{\pi(\mu-\nu)}{2} \right)J_{\nu}(rx) + \sin \left(\frac{\pi(\mu-\nu)}{2} \right)Y_{\nu}(rx) \right) \, \mathrm dx = I_{\mu}(\beta \alpha) K_{\nu}(r \alpha) $$ where $r \ge \beta >0$, and $\mu$ and $\nu$ are nonnegative real parameters such that $\mu > \nu -2$.
As in my previous answer, we can exploit properties of the Hankel function of the first kind.
Let $H_{0}^{(1)}(z)$ be the Hankel function of first kind of order zero defined as $$H_{0}^{(1)}(z) = J_{0}(z) + i Y_{0}(z), $$ where $Y_{0}(z)$ is the Bessel function of the second kind of order zero.
The principal branch of $H_{0}^{(1)}(z)$ has a branch cut on the negative real axis.
Since $$Y_{0}(xe^{i \pi})= Y_{0}(x) + 2i J_{0}(x), \quad x >0,$$ it follows that
$$H_{0}^{(1)}(xe^{i \pi}) = - J_{0}(x) + iY_{0}(x), \quad x >0. $$
And since $\frac{x}{x^{2}+\alpha^{2}}$ is an odd function, we have $$\int_{0}^{\infty} \frac{x}{x^{2}+\alpha^{2}} \, J_{0}(x) J_{0}(rx) \, \mathrm dx = \frac{1}{2} \, \Re \int_{-\infty}^{\infty} \frac{x}{x^{2}+\alpha^{2}} \, H_{0}^{(1)} (x) J_{0}(rx) \, \mathrm dx,$$ where the integration form $-\infty$ to $0$ is done on the upper side of the branch cut.
Now let's integrate the function $$f(z) = \frac{z}{z^{2}+\alpha^{2}} \, H_{0}^{(1)}(z) J_{0}(rz), \quad 0 < r \le 1 \, , $$ around a contour consisting of the upper side of the branch cut from $-R$ to $-\epsilon$, a small clockwise- oriented semicircle about the origin, the real axis from $\epsilon$ to $R$, and the upper half of the circle $|z|=R$.
Since $\lim_{z \to 0} f(z) = 0$, the contribution from the small semicircle about the origin vanishes as $\epsilon \to 0$.
And as $|z| \to \infty$ in the upper half plane, $|f(z)|$ is asymptotic to $$\frac{1}{\pi \sqrt{r}} \frac{e^{(r-1)\Im(z)}}{|z|^{2}}. $$
(See here.)
Since $0 < r \le 1$, the integral vanishes on the upper half of the circle $|z|=R$ as $R \to \infty$ (by the estimation lemma), and we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{x}{x^{2}+\alpha^{2}} \, H_{0}^{(1)} (x) J_{0}(rx) \, \mathrm dx &= 2 \pi i \operatorname*{Res}_{z=i \alpha} f(z) \\ &= 2 \pi i \lim_{z \to i \alpha} \frac{z}{z+i \alpha} \, H_{0}^{(1)}(z) J_{0}(rz) \\ &= i \pi \, H_{0}^{(1)}(i \alpha)J_{0}(i r \alpha) \\ &\overset{(\spadesuit)}{=} i \pi \left(\frac{2K_{0}(\alpha)}{i \pi} \right) I_{0}(r \alpha) \\&= 2 K_{0}(\alpha) I_{0}(r \alpha). \end{align}$$
Equating the real parts on both sides of the equation, we have $$\int_{0}^{\infty} \frac{x}{x^{2}+ \alpha^{2}} \, J_{0}(x) J_{0}(rx) \, \mathrm dx = K_{0}(\alpha) I_{0}(r \alpha), \quad 0 < r \le 1. $$
This result also holds for $r=0$.
$\spadesuit$ https://dlmf.nist.gov/10.27#E8
To show that $$\int_{0}^{\infty} \frac{x}{x^{2}+ \alpha^{2}} \, J_{0}(x) J_{0}(rx) \, \mathrm dx = I_{0}(\alpha) K_{0}(r \alpha), \quad r \ge 1, $$ integrate $$g(z) = \frac{z}{z^{2}+\alpha^{2}} \, J_{0}(z) H_{0}^{(1)}(rz) $$ around the same contour.