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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

Sakup2485
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    Please show your solution using the Mellin transform in the $xK_0^4$ case as it might be a possible approach to evaluate the generalised integrals, and also provides context for your question. – TheSimpliFire Jul 26 '21 at 08:47
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    The Mellin transform still works for $x^3,x^5,x^7\cdots$, and in general, there exists a fifth order linear recurrence to compute $\int_0^\infty x^{2n+1}K_0(x)^4 dx$. – pisco Jul 26 '21 at 09:34
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    Thanks for linking the paper. Theorem 3.2 gives an explicit closed form for the coefficient of $\zeta(3)$. A closed-form expression for the "constant" term (independent of $\zeta(3)$) remains open. Again, please include more detail (either from your attempt or from the paper) in your question. – TheSimpliFire Jul 27 '21 at 08:17

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