Jtdylktuy

Assume I have a diagonal matrix $L$ of size $n$. I want to compute the following integral:
$$I_n(L) equiv int_{(mathbb{S}^{n-1})^2} mathrm{d}sigma(x) mathrm{d}sigma(x') exp[n x^top L , x']$$
In which $mathbb{S}^{n-1}$ is the unit sphere in $mathbb{R}^n$, and $ mathrm{d}sigma$ is the usual measure on $mathbb{S}^{n-1}$, with $int_{mathbb{S}^{n-1}} mathrm{d}sigma(x) = 2 pi^{n/2}/Gamma(n/2)$.
I am looking for either a general form of $I_n(L)$ or simply an asymptotic of $frac{1}{n}log I_n(L)$ as $n to infty$.

I tried but could not find such results. Any help would be appreciated. Thanks !

There is a way to get an infinite series using the expansion of $exp$, however, it may be more efficient to compute the result using quadrature. The steps below first expand $exp$, and use the multinomial theorem to get a polynomial expansion for the integrand. Taking the integral inside the sum to the monomials we can evaluate the integral directly to obtain an analytic (closed form) expression for the integral.

For clarity, first denote the integral by $I$, i.e.
begin{align}
I := int_{S_n}int_{S_n}exp(n textbf{x}^T L textbf{y}) dsigma_n dsigma_n
end{align}

Power series for exp

If we first consider the expansion of the function argument in the integral,
begin{align}
exp(n textbf{x}^T L textbf{y}) =& sum_{k=0}^infty frac{(ntextbf{x}^T L textbf{y})^k}{k!} \
=& sum_{k=0}^infty frac{left(n sum_{r=1}^M x_r L_{rr}y_r right)^k}{k!} tag{1}
end{align}

It is well known that the series converges uniformly, and the spherical integral is linear (see [1]), which means we can take the integrals inside the summation,

begin{align}
I = sum_{k=0}^infty frac{n^k}{k!} int_{S_n}int_{S_n}left( sum_{r=1}^M x_r L_{rr}y_r right)^kdsigma_n dsigma_n tag{2}
end{align}

Use of the multinomial theorem

Using the multinomial theorem on the inner bracket gives,
begin{align}
left( sum_{r=1}^M x_r L_{rr}y_r right)^k =& sum_{j_{1} + ldots + j_{M} = k} frac{k!}{j_{1}!ldots j_{M}!} prod_{t=1}^M (L_{tt} y_t x_t)^{j_t} tag{3}
end{align}

We can then put (3) into (2) and distribute the integrals further over the addition to give,

begin{align}
I = sum_{k=0}^infty frac{n^k}{k!}sum_{j_{1} + ldots + j_{M} = k} rho_k int_{S_n}int_{S_n} prod_{t=1}^M (L_{tt} y_t x_t)^{j_{t}}dsigma_n dsigma_n tag{4}
end{align}
where $rho_k = frac{k!}{j_{1}!ldots j_{M}!} $.

Integrating a monomial

The spherical integral of a monomial is discussed in detail in [2], but the main result gives,
begin{align}
int_{S_n}int_{S_n} prod_{t=1}^M (L_{tt} y_t x_t)^{j_{t}}dsigma_n dsigma_n = left{ begin{array}{rl} left( 2frac{prod_{t=1}^MGamma(q_{t})}{Gamma(sum_{t=1}^M q_t)} right)^2 prod_{t=1}^M L_{tt}^{j_{t}} & : j_{t} mbox{ all even} \ 0& : mbox{otherwise} end{array} right .tag{5}
end{align}

Where $q_{t} = frac{1}{2} (j_{t} + 1)$. Putting (5) into (4) then gives,
begin{align}
I =& 4 sum_{k=0}^infty frac{n^k}{k!}sum_{j_{1} + ldots + j_{M} = k \ forall t mod(j_{t},2) = 0 } rho_k left(frac{prod_{t=1}^MGamma(q_{t})}{Gamma(sum_{t=1}^M q_{t})} right)^2 prod_{t=1}^M L_{tt}^{j_{t}} tag{6}
end{align}

Simplification

Since the terms are non-zero only when all of the the $j_t$, and hence $k$, are even; after some cancellations, the sum can be written,
begin{align}
I =& 4 sum_{k=0}^infty sum_{j_{1} + ldots + j_{M} = k} frac{n^{2k}}{prod_{t=1}^M (2j_{t})!}left(frac{prod_{t=1}^MGamma(j_{t} + frac{1}{2})}{Gamma(frac{M}{2} + sum_{t=1}^M j_{t})} right)^2 prod_{t=1}^M L_{tt}^{2j_{t}} \
=& 4 sum_{k=0}^infty frac{n^{2k}}{Gamma(frac{M}{2} + k)^2} sum_{j_{1} + ldots + j_{M} = k}prod_{t=1}^M frac{left(Gamma(j_{t} + frac{1}{2}) L_{tt}^{j_{t}}right)^2}{ (2j_{t})!}tag{7}
end{align}

[1] Baker, John A., Integration over spheres and the divergence theorem for balls, Am. Math. Mon. 104, No. 1, 36-47 (1997). ZBL0877.26008.

[2] Folland, Gerald B., How to integrate a polynomial over a sphere, Am. Math. Mon. 108, No. 5, 446-448 (2001). ZBL1046.26503.