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:


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.

  • $\begingroup$ Integral definition of the Bessel function seems like a way to go. $$I_0(y)=\frac{1}{\pi} \int_0^\pi e^{y \cos t}dt$$ $\endgroup$
    – Yuriy S
    Jan 23, 2017 at 8:57
  • $\begingroup$ For example $$I_0 \left(\frac{x}{3} \right)^3=\frac{1}{\pi^3} \int_0^\pi \int_0^\pi \int_0^\pi e^{\frac{x}{3} (\cos t+\cos u+\cos v)}dtdudv$$ Now you can also see the relation to $(3)$ $\endgroup$
    – Yuriy S
    Jan 23, 2017 at 9:03
  • 2
    $\begingroup$ Just to be clear $$\int_0^\infty e^{-x}I_0\left(\frac{x}{3}\right)^3\;dx=\frac{1}{\pi^3}\int_0^\infty \int_0^\pi \int_0^\pi \int_0^\pi e^{-\frac{x}{3} (3-\cos t-\cos u-\cos v)}dtdudvdx=$$ $$=\frac{3}{\pi^3} \int_0^\pi \int_0^\pi \int_0^\pi \frac{dt~du~dv}{3-\cos t-\cos u-\cos v}$$ $\endgroup$
    – Yuriy S
    Jan 23, 2017 at 9:13
  • $\begingroup$ nice integral (+1) $\endgroup$
    – tired
    Jan 23, 2017 at 17:33
  • $\begingroup$ Another form of the integral $$\int_0^\infty e^{-x}I_0\left(\frac{x}{3}\right)^3\;dx=\frac{12}{\pi^3} \int_0^1 \int_0^1 \int_0^1 \frac{dp~dq~ds}{(p^2+q^2+s^2)\sqrt{(1-p^2)(1-q^2)(1-s^2)}}$$ $\endgroup$
    – Yuriy S
    Jan 23, 2017 at 19:09

2 Answers 2


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.

  • $\begingroup$ Nice. +1. I like this way of seeing the connection with the elliptic integral. Hopefully the singular value connection can become more obvious too. $\endgroup$
    – Anon
    Jan 25, 2017 at 22:24

I am having hesitations whether I should be writing this post because I will definitely not derive the closed form in question. Yet, after having glanced through all the material linked to this question I guess I am able to contribute more than all what I have seen so far. We start from the formula (3) since the route from (1) to (3) is fairly simple and has been presented in the comments already. \begin{eqnarray} u(3) &=& \int\limits_{[0,\pi]^3} \frac{1}{3-\cos(t)-\cos(u)-\cos(v)} dt du dv\\ &=&-\pi \int\limits_{[0,\pi]^2} \frac{1}{\sqrt{-4+\cos(t)+\cos(u)}\sqrt{-2+\cos(t)+\cos(u)}} dt du\\ &\underbrace{=}_{v=\tan(u/2)}& 2\pi \int\limits_{[0,\pi] \times [0,\infty)} \frac{1}{\sqrt{(-3-5 v^2+(1+v^2) \cos(t))(-1-3 v^2 +(1+v^2) \cos(t))}}dt dv\\ &=& 2\pi \int\limits_{[0,\pi] \times [0,\infty)} \frac{1}{\sqrt{5-6 \cos(t)+\cos(t)^2}}\cdot \frac{1}{\sqrt{(1+v^2)(1+v^2 \frac{(\cos(t)-3)^2}{(\cos(t)-5)(\cos(t)-1)}}} dt dv\\ &=& 2\pi \int\limits_{[0,\pi]} \frac{K(-\frac{4}{5-6 \cos(t)+\cos(t)^2})}{\sqrt{5-6 \cos(t)+\cos(t)^2}} dt\\ &\underbrace{=}_{v=\tan(t/2)}& 2\pi \int_{[0,\infty)} \frac{K(-\frac{(1+v^2)^2}{v^2(2+3 v^2)})}{v\sqrt{2+3 v^2}} dv\\ &\underbrace{=}_{z=-\frac{(1+v^2)^2}{v^2(2+3 v^2)}}& \frac{\pi}{2} \int\limits_{(-\infty,-1/3)} \sqrt{\frac{-1+3 z-3\sqrt{z(-1+z)}}{(1+3z)(-1+z)z}} K(z) dz\\ &=& \frac{\pi}{2} \int\limits_{(-\infty,-1/3)} \sqrt{\frac{-1+3 z-3\sqrt{z(-1+z)}}{(1+3z)(-1+z)z}} \frac{1}{\sqrt{1-z}}K(\frac{z}{z-1}) dz\\ &\underbrace{=}_{y=z/(z-1)}&\frac{\imath \pi}{2} \int\limits_{(1/4,1)} \frac{K(y)}{\sqrt{y(1-3 \sqrt{y}+2 y)}}dy\\ &\underbrace{=}_{y\leftarrow \sqrt{y}}& \imath \pi \int\limits_{(1/2,1)} \frac{K(y^2)}{\sqrt{(1-y)(1-2 y)}} dy\\ &\underbrace{=}_{y\leftarrow 2y-1}& \frac{\pi}{\sqrt{2}} \int\limits_{(0,1)} \frac{K(\frac{(1+y)^2}{4})}{\sqrt{(1-y)y}} dy\\ &\underbrace{=}_{y\leftarrow \sqrt{y}}&\pi \sqrt{2} \int\limits_{(0,1)}\frac{K(\frac{(1+y^2)^2}{4})}{\sqrt{1-y^2}} dy\\ &\underbrace{=}_{y=\sin(\phi)}& \int\limits_{(0,\pi/2)} K\left( \frac{(1+\sin(\phi)^2)^2}{4}\right) d\phi\\ &\underbrace{=}_{\phi=\phi/2}& \frac{\pi \sqrt{2}}{4} \int\limits_{[-\pi,\pi]} K\left( \frac{19}{32} - \frac{3}{8} \cos(\phi)+\frac{1}{32} \cos(2 \phi)\right) d\phi\\ &\underbrace{=}_{z=\exp(\imath \phi)}& \frac{\pi \sqrt{2}}{4} \oint\limits_{|z|=1}K\left(\frac{19}{32} - \frac{3}{8}\frac{z+1/z}{2}+\frac{1}{32} \frac{z^2+1/z^2}{2} \right) \frac{dz}{\imath z} \end{eqnarray}

I guess all the steps are pretty clear. The last expression lends itself for Cauchy-residue theorem computations. However, for the time being my knowledge of the singularities of the elliptic function is not sufficient to be able to complete this computation.


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