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The Funk-Hecke formula, in its simplest form, says that for $f$, a bounded measurable function on $[-1,1]$, and $y\in S_{n}$, one has $$ \int_{S_{n}}f(<x,y>)d\sigma_{n}(x)=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{-1}^{1}f(t)(1-t^{2})^{n/2-1}dt, $$$$ \int_{S_{n}}f( \langle x,y \rangle)d\sigma_{n}(x)=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{-1}^{1}f(t)(1-t^{2})^{n/2-1}dt, $$ where $S_{n}$ is the unit sphere in $\mathbb{R}^{n+1}$, $\sigma_{n}$ is the unnormalized surface area measure on $S_{n}$, and $<x,y>$$\langle x,y \rangle$ denotes the euclideanEuclidean scalar product in $\mathbb{R}^{n+1}$.

A complex version of the Funk-Hecke formula, in $\mathbb{C}^{n}$, with the complex scalar product, has been derived in [1], Proposition 5.2, and probably more references can be found. The analog of the previous formula reads $$ \int_{S_{2n-1}}f(<x,y>)d\sigma_{2n-1}(x)=\frac{2\pi^{n-1}}{(n-2)!} \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t(1-t^{2})^{n-2}f(te^{i\theta})dtd\theta. $$$$ \int_{S_{2n-1}}f( \langle x,y \rangle)d\sigma_{2n-1}(x)=\frac{2\pi^{n-1}}{(n-2)!} \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t(1-t^{2})^{n-2}f(te^{i\theta})dtd\theta. $$ Hence, up to a multiplicative constant, your integral rewrites as $$ \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t^{a+1}(1-t^{2})^{n-2}e^{bte^{i\theta}}e^{ia\theta}dtd\theta. $$ The integral with respect to $t$ admits a reasonably simple, explicit expression in terms of hypergeometric functions (which Mathematica can compute), but I doubt that the integral with respect to $\theta$ is expressible in terms of some special functions.

[1] R.L. Frank, E.H. Lieb, Sharp constants in several inequalities on the Heisenberg group. Ann. of Math. (2) 176 (2012), 349-381.

The Funk-Hecke formula, in its simplest form, says that for $f$, a bounded measurable function on $[-1,1]$, and $y\in S_{n}$, one has $$ \int_{S_{n}}f(<x,y>)d\sigma_{n}(x)=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{-1}^{1}f(t)(1-t^{2})^{n/2-1}dt, $$ where $S_{n}$ is the unit sphere in $\mathbb{R}^{n+1}$, $\sigma_{n}$ is the unnormalized surface area measure on $S_{n}$, and $<x,y>$ denotes the euclidean scalar product in $\mathbb{R}^{n+1}$.

A complex version of the Funk-Hecke formula, in $\mathbb{C}^{n}$, with the complex scalar product, has been derived in [1], Proposition 5.2, and probably more references can be found. The analog of the previous formula reads $$ \int_{S_{2n-1}}f(<x,y>)d\sigma_{2n-1}(x)=\frac{2\pi^{n-1}}{(n-2)!} \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t(1-t^{2})^{n-2}f(te^{i\theta})dtd\theta. $$ Hence, up to a multiplicative constant, your integral rewrites as $$ \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t^{a+1}(1-t^{2})^{n-2}e^{bte^{i\theta}}e^{ia\theta}dtd\theta. $$ The integral with respect to $t$ admits a reasonably simple, explicit expression in terms of hypergeometric functions (which Mathematica can compute), I doubt that the integral with respect to $\theta$ is expressible in terms of some special functions.

[1] R.L. Frank, E.H. Lieb, Sharp constants in several inequalities on the Heisenberg group. Ann. of Math. (2) 176 (2012), 349-381.

The Funk-Hecke formula, in its simplest form, says that for $f$, a bounded measurable function on $[-1,1]$, and $y\in S_{n}$, one has $$ \int_{S_{n}}f( \langle x,y \rangle)d\sigma_{n}(x)=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{-1}^{1}f(t)(1-t^{2})^{n/2-1}dt, $$ where $S_{n}$ is the unit sphere in $\mathbb{R}^{n+1}$, $\sigma_{n}$ is the unnormalized surface area measure on $S_{n}$, and $\langle x,y \rangle$ denotes the Euclidean scalar product in $\mathbb{R}^{n+1}$.

A complex version of the Funk-Hecke formula, in $\mathbb{C}^{n}$, with the complex scalar product, has been derived in [1], Proposition 5.2, and probably more references can be found. The analog of the previous formula reads $$ \int_{S_{2n-1}}f( \langle x,y \rangle)d\sigma_{2n-1}(x)=\frac{2\pi^{n-1}}{(n-2)!} \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t(1-t^{2})^{n-2}f(te^{i\theta})dtd\theta. $$ Hence, up to a multiplicative constant, your integral rewrites as $$ \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t^{a+1}(1-t^{2})^{n-2}e^{bte^{i\theta}}e^{ia\theta}dtd\theta. $$ The integral with respect to $t$ admits a reasonably simple, explicit expression in terms of hypergeometric functions (which Mathematica can compute), but I doubt that the integral with respect to $\theta$ is expressible in terms of some special functions.

[1] R.L. Frank, E.H. Lieb, Sharp constants in several inequalities on the Heisenberg group. Ann. of Math. (2) 176 (2012), 349-381.

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created Nov 9 at 21:53

user111

The Funk-Hecke formula, in its simplest form, says that for $f$, a bounded measurable function on $[-1,1]$, and $y\in S_{n}$, one has $$ \int_{S_{n}}f(<x,y>)d\sigma_{n}(x)=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{-1}^{1}f(t)(1-t^{2})^{n/2-1}dt, $$ where $S_{n}$ is the unit sphere in $\mathbb{R}^{n+1}$, $\sigma_{n}$ is the unnormalized surface area measure on $S_{n}$, and $<x,y>$ denotes the euclidean scalar product in $\mathbb{R}^{n+1}$.

A complex version of the Funk-Hecke formula, in $\mathbb{C}^{n}$, with the complex scalar product, has been derived in [1], Proposition 5.2, and probably more references can be found. The analog of the previous formula reads $$ \int_{S_{2n-1}}f(<x,y>)d\sigma_{2n-1}(x)=\frac{2\pi^{n-1}}{(n-2)!} \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t(1-t^{2})^{n-2}f(te^{i\theta})dtd\theta. $$ Hence, up to a multiplicative constant, your integral rewrites as $$ \int_{t=0}^{1}\int_{\theta=-\pi}^{\pi}t^{a+1}(1-t^{2})^{n-2}e^{bte^{i\theta}}e^{ia\theta}dtd\theta. $$ The integral with respect to $t$ admits a reasonably simple, explicit expression in terms of hypergeometric functions (which Mathematica can compute), I doubt that the integral with respect to $\theta$ is expressible in terms of some special functions.

[1] R.L. Frank, E.H. Lieb, Sharp constants in several inequalities on the Heisenberg group. Ann. of Math. (2) 176 (2012), 349-381.