Let $I_\nu(x)$ be the modified Bessel functions of first kind with order$\text{ }\nu$, $K_\nu(x)$ be the modified Bessel functions of second kind with order$\text{ }\nu$.

Prerequisite Information:

The integral $$\int_{0}^{\infty} x^4K_0(x)K_1(x)^3 \ln(xK_1(x))^2\text{d}x=\frac{1}{32}$$ can be shown as follows:

Note that $$\frac{\text{d}}{\mathrm{d}x}\left ( -\frac{x^\alpha}{\alpha} K_1(x)^\alpha \right ) =x^\alpha K_0(x)K_1(x)^{\alpha-1}.$$ Therefore,$$\int_{0}^{\infty}x^\alpha K_0(x)K_1(x)^{\alpha-1}\text{d}x =\frac{1}{\alpha},\qquad{\Re(\alpha)>0} .$$ And the equality immediately follows by differentiating the expression.

There are also some integral identities involving Bessel functions, but not (quite) trivial. These integrals had studied in arXiv:0801.0891. For example, $$\begin{aligned} &\int_{0}^{\infty}K_0(x)^3\text{d}x=\frac{3\Gamma\left ( \frac{1}{3} \right )^6 }{32\pi\cdot2^{2/3}} ,\\ &\int_{0}^{\infty}xK_0(x)^4\text{d}x=\frac{7}{8}\zeta(3) ,\\ &\int_{0}^{\infty}xI_0(x)K_0(x)^2\text{d}x= \frac{\pi}{3\sqrt{3} },\\ &\int_{0}^{\infty}xI_0(x)K_0(x)^3\text{d}x=\frac{\pi^2}{16} . \end{aligned}$$ In this paper, the authors determine some relations among the moments. For example, $$ \int_{0}^{\infty}K_0(x)^4\text{d}x =\pi^2\int_{0}^{\infty}K_0(x)^2I_0(x)^2\mathrm{d}x. $$ These relations can be generalized in many ways. Using contour integration, we conclude that $$ \int_{0}^{\infty} x^3 K_0(x)^5I_0(x)\left ( \pi^2I_0(x)^2-K_0(x)^2 \right ) \text{d}x=\frac{\pi^4}{128}. $$ (Only one example.)
Moreover, $$ \int_{0}^{\infty} x^{2k+1} K_0(x)^5I_0(x)\left ( \pi^2I_0(x)^2-K_0(x)^2 \right ) \text{d}x= \begin{cases} 0 & k=0, \\ a_k\cdot\pi^4 & k\in\mathbb{Z}^{+}. \end{cases} $$ Where $a_k$ is always a rational number. And we are able to compute $$ \int_{0}^{\infty} xI_0(\alpha x) K_0(x)^5I_0(x)\left ( \pi^2I_0(x)^2-K_0(x)^2 \right ) \text{d}x $$ by expanding the $I_0(\alpha x)$ into Maclaurin series. Another simple identity is given by $$ \int_{0}^{\infty}x^7K_0(x)K_1(x)^2K_2(x)\text{d}x =\frac{1}{3}. $$


I am trying to find more results but failed. Can we find the closed-forms of other moments such as $\int_{0}^{\infty}K_0(x)^5\text{d}x, \int_{0}^{\infty}K_0(x)I_0(x)J_0(x)Y_0(x)\text{d}x$? Any idea would be much appreciated.

Maybe interests: Two integrals (both are easy to check): $$\begin{aligned} &\int_{0}^{\infty} \frac{x^2}{\alpha^2+x^2}K_0(x)^2\text{d}x =\frac{\pi^2}{4}-\frac{\pi^3}{8}\alpha \left ( J_0(\alpha)^2+Y_0(\alpha)^2 \right ), \\ &\int_{0}^{\infty}K_0(x)^2\cos(\alpha x)\mathrm{d}x =\frac{\pi}{\sqrt{4+\alpha^2} }K\left ( \frac{\alpha}{\sqrt{4+\alpha^2} } \right ). \end{aligned}$$ Where $K(x)=\frac\pi2{}_2F_1\left(\frac12,\frac12;1;x^2\right)$ and ${}_2F_1$ is Gauss hypergeometric function.


1 Answer 1


I don't have a closed-form expression for $$\int_{0}^{\infty} J_{0}(x) Y_{0}(x) I_{0}(x) K_{0}(x) \, \mathrm dx,$$ but the integral can be shown to have the same value as two other integrals involving the product of Bessel functions.

Consider the complex function $$f(z) = H_{0}^{(1)}(z) Y_{0}(z) I_{0}(z) K_{0}(z), $$ where $H_{0}^{(1)}(z) $ is the Hankel function of the first kind of order zero defined as $$H_{0}^{(1)}(z)= J_{0}(x)+ i Y_{0}(x). $$

The function $f(z)$ is analytic in the right half-plane.

And in the first quadrant of the complex plane (where $0 < \arg(z) < \frac{\pi}{2}$), $$ f(z) \sim \frac{i }{2 \pi}\frac{1}{z^{2}} \ \text{as} \ |z| \to \infty. $$

Therefore, we can integrate around a closed quarter circle in the first quadrant (indented at the origin) and conclude that $$\int_{0}^{\infty}H_{0}^{(1)}(x) Y_{0}(x) I_{0}(x) K_{0}(x) \, \mathrm dx - i \int_{0}^{\infty} H_{0}^{(1)}(ix) Y_{0}(ix) I_{0}(ix) K_{0}(ix) \, \mathrm dx =0,$$ where $$\small H_{0}^{(1)}(ix) Y_{0}(ix) I_{0}(ix) K_{0}(ix) = \frac{2}{i \pi} \, K_{0}(x) \left(-\frac{2}{\pi}K_{0}(x)+ i I_{0}(x) \right) J_{0}(x) \left(- \frac{\pi}{2} Y_{0}(x) - \frac{i \pi}{2} J_{0}(x) \right). $$

Equating the real parts on both sides of equation, we get $$\int_{0}^{\infty} J_{0}(x) Y_{0}(x) I_{0}(x) K_{0}(x) \, \mathrm dx = \int_{0}^{\infty} \left(J_{0}(x)^{2} I_{0}(x)K_{0}(x) + \frac{2}{\pi} \, J_{0}(x) Y_{0}(x) K_{0}(x)^{2} \right) \, \mathrm dx. $$

And equating the imaginary parts on both sides of the equation, we get $$\int_{0}^{\infty} J_{0}(x) Y_{0}(x) I_{0}(x) K_{0}(x) \, \mathrm dx = \int_{0}^{\infty} \left( \frac{2}{\pi} \, J_{0}(x)^{2}K_{0}(x)^{2}- Y_{0}(x)^{2} I_{0}(x) K_{0}(x) \right) \, \mathrm dx.$$

Also, in addition to the identity $$\int_{0}^{\infty} K_{0}(x)^{4} \, \mathrm dx = \pi^{2} \int_{0}^{\infty} K_{0}(x)^{2} I_{0}(x)^{2} \, \mathrm dx,$$ there is the identity $$\int_{0}^{\infty} J_{\alpha}(x)^{4} \, \mathrm dx = \int_{0}^{\infty} Y_{\alpha}(x)^{2}J_{\alpha}(x)^{2} \, \mathrm dx, \quad \alpha \ge 0. \tag{1}$$

I assume $(1)$ is a known identity.

The following is a way to proof $(1)$ using contour integration.

In the right half-plane, the function $$f(z) = K_{\alpha}(z)^{2} I_{\alpha}(z)^{2} $$ is analytic and asymptotic to $\frac{1}{4z^{2}}$ as $|z| \to \infty$. (See here.)

So by integrating $f(z)$ around a closed quarter-circle in the first quadrant (that is indented at the origin), we get

$$\int_{0}^{\infty} K_{\alpha}(x)^{2}I_{\alpha}(x)^{2} \, \mathrm dx - \int_{0}^{\infty}K_{\alpha}(ix)^{2} I_{\alpha}(ix)^{2} \, i \, \mathrm dx = 0. $$

But $$ \begin{align} I_{\alpha}(ix) &= \sum_{m=0}^{\infty} \frac{1}{m!\Gamma(m+\alpha +1)} \left(\frac{ix}{2} \right)^{2m+ \alpha} \\ &= e^{i \alpha \pi /2 } \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!\Gamma(m+\alpha +1)} \left(\frac{x}{2} \right)^{2m+ \alpha} \\ &= e^{i \alpha \pi /2}J_{\alpha}(x), \end{align}$$

and $$ \begin{align} K_{\alpha}(ix) &= \frac{\pi}{2} \frac{I_{-\alpha}(ix) - I_{\alpha}(ix)}{\sin(\alpha \pi)} \\ &= \frac{\pi}{2} \frac{e^{-i \alpha \pi /2} J_{-\alpha}(x) - e^{i \alpha \pi /2} J_{\alpha}(x)}{\sin(\alpha \pi)} \\ & =\frac{\pi e^{- i \alpha \pi /2} }{2} \frac{J_{-\alpha}(x) - e^{i \alpha \pi } J_{\alpha}(x)}{\sin(\alpha \pi)} \\ &= \frac{\pi e^{- i \alpha \pi /2} }{2} \frac{J_{-\alpha}(x) -\frac{1}{2} \left(e^{i \alpha \pi}+e^{- i \alpha \pi }\right)J_{\alpha}(x)- \frac{1}{2}\left( e^{i \alpha \pi } -e^{-i \alpha \pi } \right)J_{\alpha}(x)}{\sin(\alpha \pi) } \\ &= \frac{\pi e^{- i \alpha \pi /2} }{2} \frac{J_{-\alpha}(x) -\cos(\alpha \pi) J_{\alpha}(x)- i \sin(\alpha \pi) J_{\alpha}(x)}{\sin(\alpha \pi) } \\ &= -\frac{\pi e^{- i \alpha \pi /2} }{2} \left( Y_{\alpha}(x)+ i J_{\alpha}(x) \right). \end{align}$$

Therefore, we have $$\int_{0}^{\infty} K_{\alpha}(x)^{2}I_{\alpha}(x)^{2} \, \mathrm dx - \frac{i \pi^{2}}{4} \int_{0}^{\infty} \left(Y_{\alpha}(x)+iJ_{\alpha}(x) \right)^{2} J_{\alpha}(x)^{2} \, \mathrm dx = 0. $$

Equating the imaginary parts on both sides of the equation, we get $$\int_{0}^{\infty} \left( Y_{\alpha}(x)^{2}J_{\alpha}(x)^{2}- J_{\alpha}(x)^{4} \right) \, \mathrm dx =0. $$

Also, by equating the real parts on both sides of the equation, we get $$\int_{0}^{\infty} K_{\alpha}(x)^{2} I_{\alpha}(x)^{2} \, \mathrm dx + \frac{\pi^{2}}{2} \int_{0}^{\infty} Y_{\alpha}(x) J_{\alpha}(x)^{3} \, \mathrm dx =0. $$


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