TITLE: Generalized integrals for Bessel Moments $\int_{0}^{\infty} x^4K_0(x)K_1(x)^3 \ln(xK_1(x))^2\text{d}x=\frac{1}{32}$
QUESTION [7 upvotes]: 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
$$
\int_{0}^{\infty}x^7K_0(x)K_1(x)^2K_2(x)\text{d}x
=\frac{1}{3}.
$$

Problem:
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.

REPLY [2 votes]: One may derive an integral relation:

$$I_1=\int_{-1}^{1} \int_{-1}^{1} \int_{0}^{1} 
\frac{x}{\sqrt{1-x^2}\sqrt{1-y^2} \sqrt{1-z^2}  } 
\frac{1+x(4+y+z)}{(1+x(3+y+z))^{3/2}(1+x(5+y+z))^{3/2}} 
\text{d}x\text{d}y\text{d}z$$
$$
I_2=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 
\frac{1}{\sqrt{1-x^2}\sqrt{1-y^2} \sqrt{1-z^2}  } 
\frac{xy+yz+xz+4xyz}{(xy+yz+xz+3xyz)^{3/2}(xy+yz+xz+5xyz)^{3/2}} 
\text{d}x\text{d}y\text{d}z
$$
Then$$I_1-I_2=\frac{\pi^2}{16}.$$

They are a little bit like Watson's triple integrals:
$$\begin{aligned}
W_1 & = \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} 
\frac{1}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}} 
\frac{1}{1-xyz}\text{d}x\text{d}y\text{d}z\\
W_2&=\int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} 
\frac{1}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}} 
\frac{1}{3-xy-yz-xz}\text{d}x\text{d}y\text{d}z\\
W_3&=\int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} 
\frac{1}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}   } 
\frac{1}{3-x-y-z}\text{d}x\text{d}y\text{d}z
\end{aligned}$$

After some calculations, I got
$$\begin{aligned}
&\int_{0}^{\infty}e^{-2x}I_0(x)K_0(x)\text{d}x  = \frac{\pi}{4}\\
&\int_{0}^{\infty}\sqrt{x}e^{-3x}K_0(x)\left ( \pi^2I_0(x)^2-\frac{1}{3}K_0(x)^2  \right ) \text{d}x
=\frac{\pi^{5/2}}{6\sqrt{2} }\\
&\int_{0}^{\infty}x^2K_0(x)^2
\left ( \frac{3\pi^2}{2} e^{-2x}I_0(x)^2-K_0(x)^2\cosh(2x)\right )
\text{d}x
=\frac{\pi^3}{32}\\
&\int_{0}^{\infty}x^9K_0(x)^7I_0(x)\left ( K_0(x)^4+\pi^4I_0(x)^4
-\frac{10\pi^2}{3}K_0(x)^2I_0(x)^2 \right ) \text{d}x=\frac{3\pi^6}{1536}
\end{aligned}$$
And a cute-looking integral with Struve L function.
$$\int_{0}^{\infty}K_0(x)(I_0(x)-\mathbf{L}_0(x))\text{d}x
=\frac{4C}{\pi}.$$
Where $C=L_{-4}(2)$ is Catalan's constant.