# Strategies for solving the infinite integral $\int_0^\infty \frac{k}{k^2+\alpha^2} \, J_0(k) J_0(kr) \, \mathrm{d}k$ where $r \ge 0$ and $\alpha > 0$

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.

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.

• Thanks for your great answer. Suppose now that we have $J_2(kr)$ instead of $J_0(kr)$ in the integrand. What I get for $r \ge 1$ is $2K_0(\alpha)I_1(\alpha r)/(\alpha r) - I_0(\alpha) K_0(\alpha r)$ but when using exemplary numerical values I do not get the same results. Any thoughts why? Commented Nov 3, 2023 at 7:48
• @preuss Then the contribution from the small semicircle about the origin doesn't vanish in the limit. I'm getting $$\int_{0}^{\infty} \frac{x}{x^{2}+\alpha^{2}} \, J_{0}(x) J_{2}(rx) \, \mathrm dx = - I_{0}(\alpha) K_{2}(r \alpha) + \frac{2}{\alpha^{2}r^{2}}, \quad r \ge 1.$$ Commented Nov 3, 2023 at 8:39
• That's brilliant. Thanks a lot for your help in all of that.Very much appreciated! Commented Nov 3, 2023 at 8:42
• Nice expression. But it looks like this is true only when $\mu \ge \nu$. Commented Nov 8, 2023 at 8:23
• @preuss Thanks for the bounty. You are correct. We need restrictions on $\mu$ and $\nu$ so that the integral on the small semicircle vanishes as $\epsilon \to 0$. Basically what we need is for $\lim_{z \to 0} \frac{z^{\color{red}{2}}}{z^{2}+\alpha^{2}} J_{\mu}(\beta z) H_{\nu}^{(1)}(rz) =0$ to hold. If $\mu$ and $\nu$ are nonnegative real values, we need $\mu > \nu -2$. Commented Nov 8, 2023 at 18:46

Using CAS like Maple give me:

$$\int_0^{\infty } \frac{k J_0(k) J_0(k r)}{k^2+\alpha ^2} \, dk=I_0(r \alpha ) K_0(\alpha ) \theta (1-r)+I_0(\alpha ) K_0(r \alpha ) \theta (-1+r)$$

where:

$$\theta (r)$$ is Heaviside theta function.

$$K_0(\alpha )$$ is modified Bessel function of the second kind.

$$I_0(\alpha )$$ is modified Bessel function of the first kind.

Maple 2023.2 code:

inttrans:-hankel(BesselJ(0, k)/(alpha^2 + k^2), k, r, 0)

• Thanks for the answer. Which Maple version do you use? Mine is 2021 and it does not know how to deal with it. Same for Mathematica. Commented Nov 2, 2023 at 17:44
• @preuss It's 2023.2 Commented Nov 2, 2023 at 17:44
• I rectify: yes, 2021 does it as well when using inttrans. Thanks for pointing out this use. It works now and i am happy. Commented Nov 2, 2023 at 17:49
• It works for me with Maple 2020.2 Commented Nov 2, 2023 at 17:50