Wolfram Alpha says:
$$
\int_{0}^{\infty} xK_0(x)^4\text{d}x
=\frac{7\zeta(3)}{8}
$$
Where $$K_0(x)
=\int_{0}^{\infty} e^{-x\cosh z}\text{d}z
$$
And I proved it by using Mellin transform.
But I also found(guess):
$$
\begin{aligned}
&\int_{0}^{\infty}x^3K_0(x)^4\text{d}x
=\frac{7\zeta(3)-6}{32} \\
&\int_{0}^{\infty}x^5K_0(x)^4\text{d}x
=\frac{49\zeta(3)-54}{128}\\
&\int_{0}^{\infty}x^7K_0(x)^4\text{d}x
=\frac{1008\zeta(3)-1184}{512}\\
&\int_{0}^{\infty} x^9K_0(x)^4\text{d}x
=\frac{42777\zeta(3)-51110}{2048}
\end{aligned}
$$
How to prove them?The Mellin transform doesn't work on $x^3,x^5,x^7...$.Any help be appreciated.
Please see:Hypergeometric Forms for Ising-Class Integrals