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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – keynes
    Commented Nov 3, 2023 at 7:48
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    $\begingroup$ @preuss Then the contribution from the small semicircle about the origin doesn't vanish in the limit. I'm getting $$\int_{0}^{\infty} \frac{x}{x^{2}+\alpha^{2}} \, J_{0}(x) J_{2}(rx) \, \mathrm dx = - I_{0}(\alpha) K_{2}(r \alpha) + \frac{2}{\alpha^{2}r^{2}}, \quad r \ge 1. $$ $\endgroup$ Commented Nov 3, 2023 at 8:39
  • $\begingroup$ That's brilliant. Thanks a lot for your help in all of that.Very much appreciated! $\endgroup$
    – keynes
    Commented Nov 3, 2023 at 8:42
  • $\begingroup$ Nice expression. But it looks like this is true only when $\mu \ge \nu$. $\endgroup$
    – keynes
    Commented Nov 8, 2023 at 8:23
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    $\begingroup$ @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 $\mu > \nu -2$. $\endgroup$ Commented Nov 8, 2023 at 18:46
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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)

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  • $\begingroup$ 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. $\endgroup$
    – keynes
    Commented Nov 2, 2023 at 17:44
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    $\begingroup$ @preuss It's 2023.2 $\endgroup$ Commented Nov 2, 2023 at 17:44
  • $\begingroup$ I rectify: yes, 2021 does it as well when using inttrans. Thanks for pointing out this use. It works now and i am happy. $\endgroup$
    – keynes
    Commented Nov 2, 2023 at 17:49
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    $\begingroup$ It works for me with Maple 2020.2 $\endgroup$
    – GEdgar
    Commented Nov 2, 2023 at 17:50

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