How to calculate an integral over the complex unit sphere

Question

We want to calculate the following integral over the complex unit sphere $S^{2n-1}$:

$$\int_{S^{2n-1}} \frac{1 }{|1 - \langle z, \zeta \rangle|^2} \, d\sigma(\zeta),$$

where $ z $ is a fixed point in the complex unit ball $ \mathbb{C}^n $ and $\zeta $ is a point on the unit sphere $ S^{2n-1} $. The notation $ \langle z, \zeta \rangle $ denotes the Hermitian inner product, defined as:

$$\langle z, \zeta \rangle = \sum_{j=1}^{n} z_j \overline{\zeta_j},$$

where $\overline{\zeta_j}$ is the complex conjugate of $\zeta_j $.

Answer 1

I shall assume that $\sigma$ is the probability measure on $S^{2n - 1}$. We write $z = r\zeta'$, where $r \in [0,1)$ and $\zeta' \in S^{2n - 1}$. From this, we see that the integral depends only on $r$, so it suffices to take $\zeta' = e_n = (0,\ldots,0,1)$. We write $\zeta = \sqrt{1 - |z|^2} (\xi,0) + z e_n$, where $z \in \mathbb{D}$, the unit disc in $\mathbb{C}$, and $\xi \in S^{2n - 3}$. Since the integrand only depends on $r$ and $z$, the integral becomes $$\frac{n - 1}{\pi} \int_{\mathbb{D}} \frac{1}{|1 - rz|^2} (1 - |z|^2)^{n - 2} \, d\mu(z),$$ where $d\mu(z)$ denotes the Lebesgue measure on $\mathbb{C}$ (see e.g. Section 1.4.5 of Rudin's book "Function Theory in the Unit Ball of $\mathbb{C}^n$"). Passing to polar coordinates, this becomes $$\frac{n - 1}{\pi} \int_{0}^{1} (1 - \rho^2)^{n - 2} \rho \int_{-\pi}^{\pi} \frac{1}{1 - 2r \rho \cos \theta + r^2 \rho^2} \, d\theta \, d\rho.$$ The antiderivative with respect to $\theta$ of the innermost integrand is $$\frac{2}{1 - r^2 \rho^2} \arctan \left( \frac{1 + r\rho}{1 - r\rho} \tan \frac{\theta}{2}\right),$$ and so the innermost integral is equal to $\frac{2\pi}{1 - r^2 \rho^2}$. Thus the original integral is equal to $$2(n - 1) \int_{0}^{1} \frac{(1 - \rho^2)^{n - 2} \rho}{1 - r^2 \rho^2} \, d\rho,$$ which takes the form $2(n - 1) (r^{-2(n - 2)} P_{n - 3}(r^2) + \frac{1}{2} r^{-2(n - 1)} (1 - r^2)^{n - 2} \log(1 - r^2))$ for some polynomial $P_{n - 3}$ of degree $n - 3$.

Answer 2

My formula in the comment for $\|z\|=1$, was obtained as follows.

First note that the integral is well defined in $[0,\infty]$, and by monotone convergence, is given by $$ \int_{S^{2n-1}}\frac{1}{|1-\langle z,\zeta\rangle|^2}\ d\sigma(\zeta) =\lim_{r\rightarrow 1^{-}} \int_{S^{2n-1}}\frac{1}{|1-r\langle z,\zeta\rangle|^2}\ d\sigma(\zeta) $$ Now for $r\in(0,1)$, one can use the convergent representation $$ \frac{1}{1-r\langle z,\zeta\rangle}=\sum_{k\ge 0} r^k \langle z,\zeta\rangle^k $$ and its conjugate, in order to write $$ \int_{S^{2n-1}}\frac{1}{|1-r\langle z,\zeta\rangle|^2}\ d\sigma(\zeta) =\int_{S^{2n-1}} \sum_{k,\ell\ge 0} r^{k+\ell} \langle z,\zeta\rangle^k \langle \zeta,z\rangle^{\ell} \ d\sigma(\zeta) $$ $$ =\sum_{k,\ell\ge 0} r^{k+\ell} \int_{S^{2n-1}} \langle z,\zeta\rangle^k \langle \zeta,z\rangle^{\ell} \ d\sigma(\zeta) $$ The analogue of the Isserlis-Wick formula for integrating monomials on the complex unit sphere is $$ \int_{S^{2n-1}} \zeta_{i_1}\cdots \zeta_{i_k}\ \overline{\zeta_{j_1}}\cdots\overline{\zeta_{j_{\ell}}}\ \ d\sigma(\zeta) =\delta_{k,\ell}\ \frac{k!}{n(n+1)\cdots(n+k-1)}\ S_{i_1,\ldots,i_k}^{j_1,\ldots,j_k} $$ where $$ S_{i_1,\ldots,i_k}^{j_1,\ldots,j_k}:=\frac{1}{k!}\ \sum_{\sigma\in\mathfrak{S}_k}\delta_{i_1,j_{\sigma(1)}}\cdots\delta_{i_k,j_{\sigma(k)}} $$ is a symmetrizer. The formula holds for all assignments of the $i$ and $j$ indices in $\{1,2,\ldots,n\}$. This is easily proved from the Isserlis-Wick theorem for the complex Gaussian $$ \frac{1}{\pi^n}\int_{\mathbb{C}^n} e^{-|\langle \zeta,\zeta\rangle|^2} \ \zeta_{i_1}\cdots \zeta_{i_k}\ \overline{\zeta_{j_1}}\cdots\overline{\zeta_{j_{\ell}}}\ \ \prod_{a=1}^{n}d({\rm Re}\ \zeta_a)\ d({\rm Im}\ \zeta_a) $$ $$ =\delta_{k,\ell}\sum_{\sigma\in\mathfrak{S}_k}\delta_{i_1,j_{\sigma(1)}}\cdots\delta_{i_k,j_{\sigma(k)}}\ . $$ by separating the radial part from the spherical part, using the homogeneity of the integrand.

As a result, $$ \int_{S^{2n-1}} \langle z,\zeta\rangle^k \langle \zeta,z\rangle^{\ell} \ d\sigma(\zeta) $$ vanishes when $k\neq \ell$, and is equal to $$ \frac{k!}{n(n+1)\cdots(n+k-1)}=\frac{1}{\binom{n+k-1}{k}}=\frac{(n-1)!}{(k+1)(k+2)\cdots(k+n-1)} $$ when $k=\ell$. We then get the convergent series representation $$ \int_{S^{2n-1}}\frac{1}{|1-r\langle z,\zeta\rangle|^2}\ d\sigma(\zeta) =(n-1)!\ \sum_{k=0}^{\infty}\frac{r^{2k}}{(k+1)(k+2)\cdots(k+n-1)} $$ After taking the limit $r\rightarrow 1$, we get $$ \int_{S^{2n-1}}\frac{1}{|1-\langle z,\zeta\rangle|^2}\ d\sigma(\zeta) =(n-1)!\ \sum_{k=0}^{\infty}\frac{1}{(k+1)(k+2)\cdots(k+n-1)} $$ which diverges for $n=1,2$, and otherwise is a telescopic sum, given the identity $$ \frac{1}{(k+1)(k+2)\cdots(k+n-1)}=\frac{1}{(n-2)} \left[ \frac{1}{(k+1)(k+2)\cdots(k+n-2)} -\frac{1}{(k+2)(k+3)\cdots(k+n-1)} \right]\ . $$ Finally, for $n\ge 3$, $$ \int_{S^{2n-1}}\frac{1}{|1-\langle z,\zeta\rangle|^2}\ d\sigma(\zeta)=\frac{n-1}{n-2}\ . $$

Now going back to the original question of computing the integral when $z$ is in the open unit ball, rather than the sphere (as I misunderstood), one has, by the above computation, $$ \int_{S^{2n-1}}\frac{1}{|1-\langle z,\zeta\rangle|^2}\ d\sigma(\zeta) =(n-1)!\ \sum_{k=0}^{\infty}\frac{r^{2k}}{(k+1)(k+2)\cdots(k+n-1)}\ , $$ when $\|z\|=r\in (0,1)$. With the telescoping sum identity above, one can then put this in the form mentioned by Peter. I don't know if one has completely explicit formulas for the result which essentially is a hypergeomtric series $$ {}_2 F_1\left[\begin{array}{c}1,1\\ n\end{array}; r^2\right]\ . $$ Note that Peter's answer is of course simpler than mine, but I wanted to advertise the Isserlis-Wick approach which is much more powerful than say Funk-Hecke, because it can also be used for integrating over several vectors $\zeta$. It also shows that what looks like a problem of multivariate calculus is really a problem of combinatorics. See these two other MO questions for similar examples:

Integration of a function over 7-sphere

Moments of Plücker coordinates on complex Grassmannian