I know that the following identity is correct, but I would love to see a derivation:
$$\int_0^\infty e^{-x}I_0\left(\frac{x}{3}\right)^3\;dx=\frac{\sqrt{6}}{32\pi^3}\Gamma\left(\frac{1}{24}\right)\Gamma\left(\frac{5}{24}\right)\Gamma\left(\frac{7}{24}\right)\Gamma\left(\frac{11}{24}\right)\tag{1}$$
where $I_0(x)$ is a modified Bessel function of the first kind. The form of the RHS makes me assume that the elliptic integral singular value $K(k_6)$ is somehow involved but I see no obvious way of transforming the LHS into an elliptic integral. My request is to see a way of evaluating this integral directly to show that it equals the RHS.
Background: This integral arises in the theory of the recurrence of random walks on a cubic lattice. Jonathan Novak elegantly shows here using generating functions that the recurrence probability of a random walk on $\mathbb{Z}^d$ is $p(d)=1-\frac{1}{u(d)}$ where:
$$u(d)=\int_0^\infty e^{-x}I_0\left(\frac{x}{d}\right)^d\;dx\tag{2}$$
The MathWorld page lists several ways of writing $u(3)$, including $(1)$, but in none of the references could I find a specific derivation of $(1)$. Most relevant discussions I found (e.g. here and here where elliptic integrals do appear) work from an alternative way of writing $u(3)$, namely:
$$u(3)=\frac{3}{2\pi^3}\int_{-\pi}^\pi\int_{-\pi}^\pi\int_{-\pi}^\pi\frac{dx\;dy\;dz}{3-\cos{x}-\cos{y}-\cos{z}}\tag{3}$$
rather than a representation like $(2)$. I believe $(3)$ comes from an alternative way of solving the random walk problem (e.g. see here) which I am not very familiar with. Accordingly, I am asking for a derivation of $(1)$ that works directly from the Bessel function integral, rather than considering other ways of calculating $u(3)$ (deriving the RHS of $(1)$ from $(3)$ would not answer my question). Although I appreciate that there may be simpler proofs of $(1)$ from the theory of random walks, I am wondering if there is a nice way of evaluating the integral in $(1)$ directly.
I think many identities are involved here, so let we outline a partial answer (for now). We have $$ I_0(z) = \sum_{n\geq 0}\frac{z^{2n}}{4^n n!^2},\qquad \mathcal{L}(I_0(z))=\frac{\mathbb{1}_{s>1}}{\sqrt{s^2-1}}\tag{1} $$ and $$ I_0(z)^2 = \sum_{n\geq 0}\frac{z^{2n}}{4^n}\sum_{k=0}^{n}\frac{1}{k!^2(n-k)!^2} = \sum_{n\geq 0}\frac{z^{2n}}{4^n n!^2}\sum_{k=0}^{n}\binom{n}{k}^2=\sum_{n\geq 0}\frac{z^{2n}\binom{2n}{n}}{4^n n!^2}\tag{2}$$ so: $$ \mathcal{L}(I_0(z)^2) =\frac{2}{\pi}\,K\!\left(\frac{2}{s}\right)\mathbb{1}_{s>2}\tag{3}$$ and the connection with the complete elliptic integral of the first kind is less obscure. If we set $$ A_n = \sum_{k=0}^{n}\frac{\binom{2k}{k}}{k!^2(n-k)!^2}\tag{4}$$ we have: $$ I_0(z)^3 = \sum_{n\geq 0}\frac{z^{2n}}{4^n}A_n \tag{5} $$ and $$\begin{eqnarray*} \int_{0}^{+\infty}I_0(z)^3 e^{-3z}\,dz = \sum_{n\geq 0}\frac{(2n)!}{4^n 3^{2n+1}}\,A_n&=&\sum_{0\leq k\leq n}\frac{(2n)!\binom{2k}{k}}{4^n 3^{2n+1} k!^2 (n-k)!^2}\\&=&\sum_{k\geq 0}\frac{(2k)!}{k!^4}\sum_{n\geq k}\frac{(2n)!}{4^n 3^{2n+1}(n-k)!^2}.\tag{6}\end{eqnarray*}$$ The definitive connection with a singular value should now be a consequence of the identity $$ K(k)^2 = \frac{\pi^2}{4}\sum_{n\geq 0}\left(\frac{(2n-1)!!}{(2n)!!}\right)^3 (2kk')^{2n} \tag{7}$$ that holds for any $0<k<\frac{1}{\sqrt{2}}$ due to the (AGM-like) invariance properties of the complete elliptic integral of the first kind.