# How to calculate an integral over the complex unit sphere

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 $$\zeta_j$$.

• Only looked at the question very quickly, so I could be wrong, but I think the integral diverges for $n=1,2$ and then is equal to $1/(n-2)\times 1/(n-2)!$ for $n>2$, if the measure $\sigma$ is normalized as a probability measure. Commented Aug 5 at 15:30
• I forgot to say, I misread the assumption on $z$ and did the computation for the case where it is on the unit sphere. Commented Aug 5 at 20:41
• The formula in the previous comment is also off, because I forgot to multiply by $(n-1)!$. See my answer below fr details. Commented Aug 5 at 21:24
• You may also use the complex version of the Funk-Hecke formula for such an integral. Commented Aug 6 at 14:14
• @user111 That is precisely what I did in my answer in moving from an integral over $S^{2n - 1}$ to an integral over $\mathbb{D}$. Commented Aug 6 at 16:16

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 $$n - 3$$.

• Thank you for your helo@Peter Humphries Commented Aug 6 at 13:59

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

• Thank you for your helo@Abdemalek Abdesselam Commented Aug 6 at 13:58