Chebyshev in $n$ dimensions

Question

The Chebyshev monomial integrals (of first and second kind) are $$ I_k = \int_{-1}^1 x^k (1-x^2)^{\mp 1/2} \,dx $$ Is anything known about their $n$-dimensional generalizations $$ I_{k_1,\dots,k_n} = \int_{S_n} \left(\prod_{i=1}^n x_i^{k_i}\right) \left(1-\sum_{i=1}^n x_i^2\right)^{\mp 1/2} \,dx $$ with $S_n=\{(x_1,\dots,x_n): \sum_{i=1}^n x_i^2 \le 1\}$?

Answer 1

Clearly $I_{k_1,\ldots,k_n}=0$ if at least one of $k_1,\ldots,k_n$ is odd. Otherwise we can go even more general. The integrand (as well as the domain of integration) is symmetric w.r.t. $x_i\leftrightarrow-x_i$; substituting $x_i^2=y_i$, $$I_{k_1,\ldots,k_n}=J_n\left(\frac{k_1+1}{2},\ldots,\frac{k_n+1}{2};1\mp\frac12\right),$$ where, for $\alpha_1,\ldots,\alpha_n,\beta>0$, $$J_n(\alpha_1,\ldots,\alpha_n;\beta):=\idotsint\limits_{\substack{0<y_1,\ldots,y_n\\y_1+\ldots+y_n<1}}y_1^{\alpha_1-1}\cdots y_n^{\alpha_n-1}(1-y_1-\ldots-y_n)^{\beta-1}\,dy_1\cdots dy_n.$$ As a repeated integral, with the "very inner" integration over $0<y_n<\bar{y}_n:=1-y_1-\ldots-y_{n-1}$: $$\int_0^{\bar{y}_n}y_n^{\alpha_n-1}(1-\bar{y}_n-y_n)^{\beta-1}\,dy_n\underset{y_n=(1-\bar{y}_n)t}{\phantom{\big[}=\phantom{\big]}}(1-\bar{y}_n)^{\alpha_n+\beta-1}\underbrace{\int_0^1 t^{\alpha_n-1}(1-t)^{\beta-1}\,dt}_{=\mathrm{B}(\alpha_n,\beta)},$$ we see that $J_n(\alpha_1,\ldots,\alpha_n;\beta)=\mathrm{B}(\alpha_n,\beta)J_{n-1}(\alpha_1,\ldots,\alpha_{n-1};\alpha_n+\beta)$; by induction, we get $$J_n(\alpha_1,\ldots,\alpha_n;\beta)=\frac{\Gamma(\alpha_1)\cdots\Gamma(\alpha_n)\cdot\Gamma(\beta)}{\Gamma(\alpha_1+\ldots+\alpha_n+\beta)}.$$ This provides an answer to your question.

Answer 2

From Folland we learn how to integrate monomials over the sphere $S_n$: $$ \int_{S_n} x^{k_1}\cdots x^{k_n}\,dx = \frac{2\prod_{i=1}^n\Gamma\left(\frac{k_i+1}{2}\right)}{\Gamma\left(\sum_{i=1}^n\frac{k_i+1}{2}\right)} $$ We can make use of this for our integral: $$ \begin{split} \int_{\mathbb{R}^n} x^{k_1}\cdots x^{k_n} \left(1 - \sum_{i=1}^n x_i^2\right)^\lambda \,dx &= \int_{S_n}\int_0^1 r^{n-1} r^{\sum{k_i}} (1-r^2)^\lambda x'^{k_1}\cdots x_n'^{k_n} \,dr\,d\sigma(x')\\ &= \int_0^1 r^{n-1} r^{\sum{k_i}} (1-r^2)^\lambda\,dr \times \int_{S_n} x'^{k_1}\cdots x_n'^{k_n} \,d\sigma(x') \end{split} $$ with $x_i' = x_i / r$. The one-dimensional integral in $r$ can be evaluated explicitly such that $$ \begin{split} I_{k_1,\dots,k_n} &= \frac{\Gamma\left(\frac{n + \sum k_i}{2}\right)\Gamma(1+\lambda)}{2\Gamma\left(\frac{n+\sum k_i}{2} + \lambda + 1\right)} \times \frac{2\prod_{i=1}^n\Gamma\left(\frac{k_i+1}{2}\right)}{\Gamma\left(\sum_{i=1}^n\frac{k_i+1}{2}\right)}\\ &= \frac{ \Gamma(1+\lambda)\prod_{i=1}^n\Gamma\left(\frac{k_i+1}{2}\right) }{ \Gamma\left(\sum\frac{k_i+1}{2} + \lambda + 1\right) }. \end{split} $$