The integral $$Z_3(\lambda_1,\lambda_2,\lambda_3)=\frac{1}{2}\int_{1}^1 I_0\left[\tfrac{1}{2}(\lambda_1\lambda_2) (1x)\right] I_0\left[\tfrac{1}{2} (\lambda_1+\lambda_2)(1+x)\right]\,e^{\lambda_3 x}\,dx,$$ with $\lambda_1,\lambda_3,\lambda_3\in\mathbb{R}$ and $I_0$ a Bessel function, arises in the evaluation of an integral over SO(3). From that derivation it follows that $Z_3(\lambda_1,\lambda_2,\lambda_3)$ is invariant upon interchange of $\lambda_1$ and $\lambda_3$. I am unable to rewrite this integral in a manifestly permutationinvariant form, can someone help me out?
2 Answers
Using the variable change $x=2t1$, the integral $Z_3(\lambda_1,\lambda_2,\lambda_3)$ can be put into the form \begin{equation} Z_3(\lambda_1,\lambda_2,\lambda_3)=\int_{0}^1 I_0\left[(\lambda_1\lambda_2) (1t)\right] e^{\lambda_3(1t)}I_0\left[(\lambda_1+\lambda_2)t\right]\,e^{\lambda_3 t}\,dt \end{equation} or, alternatively, $$ Z_3(\lambda_1,\lambda_2,\lambda_3)=Z(1)$$ where \begin{equation} Z(T)=\int_{0}^T I_0\left[(\lambda_1\lambda_2) (Tt)\right] e^{\lambda_3(Tt)}I_0\left[(\lambda_1+\lambda_2)t\right]\,e^{\lambda_3 t}\,dt \end{equation} $Z(t)$ can be considered as a convolution integral. Its Laplace transform is \begin{equation} \mathcal{L}[Z](s)=\frac{1}{\sqrt{(s+\lambda_3)^2(\lambda_1\lambda_2)^2}}\frac{1}{\sqrt{(s\lambda_3)^2(\lambda_1+\lambda_2)^2}} \end{equation} The square of the denominator of this transform is a monic polynomial which roots are $\left\lbrace \lambda_3+\lambda_1\lambda_2, \lambda_3\lambda_1+\lambda_2, \lambda_3\lambda_1\lambda_2,\lambda_3+\lambda_1+\lambda_2\right\rbrace$. This set is invariant upon interchange of the parameters. Thus, $Z_3(\lambda_1,\lambda_2,\lambda_3)$ must share the same property. If an integral form needed, $Z_3(\lambda_1,\lambda_2,\lambda_3)$ can be written as the inverse Laplace transform of this function which is permutation invariant: $$Z_3(\lambda_1,\lambda_2,\lambda_3)=\frac 1{2i\pi}\int_{ci\infty}^{c+i\infty}\frac{e^s\,ds}{\sqrt{P(s)}}$$ where $c>\max(r_1,r_2,r_3,r_4)$, $r_i$ are the roots given above and $$P(s)=s^42s^2(\lambda_1^2+\lambda_2^2+\lambda_3^2)8s\lambda_1\lambda_2\lambda_3+(\lambda_1\lambda_2\lambda_3)(\lambda_2\lambda_3\lambda_1)(\lambda_3\lambda_1\lambda_2)(\lambda_1+\lambda_2+\lambda_3) $$

$\begingroup$ The polynomial can be written in terms of the three invariants $a$, $b$, and $c$ in my tentative answer as $s^42 a s^28c s+2ba^2$, which is even more explicitly invariant. $\endgroup$– AdamJun 13, 2018 at 7:08
Using the expansion of the Bessel function $$ I_0(z)=\sum_{n\geq 0}\frac{\left(\frac{z}{2}\right)^{2n}}{(n!)^2}, $$ we can perform the integral over $x$ to obtain $$ Z_3=\sum_{n,m\geq 0}\frac{e^{\text{$\lambda $3}} \Gamma \left(m+\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right) (\text{$\lambda $1}+\text{$\lambda $2})^{2 m} (\text{$\lambda $2}\text{$\lambda $1})^{2 n} \, _1\tilde{F}_1\left(2 m+1;2 (m+n+1);2 \text{$\lambda $3}\right)}{\pi \Gamma (m+1) \Gamma (n+1)} $$ with $_1\tilde{F}_1\left(a;b;z\right)$ a regularized hypergeometric function.
Although this formulation is not very practical, it is useful to study the expansion of $Z_3$ in power of the $\lambda$'s. For instance, to order $6$, one gets, $$ 1+\frac{1}{6} \lambda _1 \lambda _2 \lambda _3+\frac{1}{6} \left(\lambda _1^2+\lambda _2^2+\lambda _3^2\right)+\frac{1}{60} \lambda _1 \lambda _2 \lambda _3\left(\lambda _1^2+\lambda _2^2+\lambda _3^2\right) +\frac{1}{120} \left(\lambda _1^4+\lambda _2^4+\lambda _3^4+4 \lambda _1^2\lambda _2^2+4 \lambda _1^2\lambda _3^2+4 \lambda _2^2 \lambda _3^2\right)+\frac{4 \left(\lambda _1^2+\lambda _2^2+\lambda _3^2\right){}^33 \left(\lambda _1^4+\lambda _2^4+\lambda _3^4\right) \left(\lambda _1^2+\lambda _2^2+\lambda _3^2\right)+24 \lambda _1^2 \lambda _2^2 \lambda _3^2}{5040}+\ldots $$ It is not too hard to convince oneself that $Z_3$ is indeed a function of $a=\lambda _1^2+\lambda _2^2+\lambda _3^2$, $b=\lambda _1^4+\lambda _2^4+\lambda _3^4$ and $ c=\lambda _1 \lambda _2 \lambda _3$ only, as expected from the fact that $Z_3$ is a generating functional of orthogonal matrices. Unfortunately, it is not clear to me what the pattern of the coefficients is, and a resummation of the series is out of reach right now.
Another possibility to compute $Z_3$ is to use the fact that it is an eigenfunction of some Laplacian (see question quoted by the OP). (Note that this can be checked with the explicit function above order by order in the$\lambda$'s.) Writing $$ Z_3=\sum_{p,m,n\geq0}c_{p,m,n} a^p b^m c^n, $$ one obtain the following recurrence relation for the coefficients $c_{p,m,n}$: $$ c_{p,m,n}=\frac{1}{6} \bigg(16 (m+1) (m+2) c_{p3,m+2,n}+(n+1) (n+2) c_{p2,m,n+2}+8 (m+1) (6 m+2 n+7) c_{p1,m+1,n}(n+1) (n+2) c_{p,m1,n+2}+96 (m+1) (m+2) c_{p,m+2,n2}+4 (p+1) (8 m+6 n+2 p+9) c_{p+1,m,n}\bigg), $$ which is unfortunately not easier to solve.