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I am trying to evaluate the following integral:

$$ \int_{\mathbb S^{d-1}} \exp \bigg(-\frac{(1+x\cdot y)^2}{\|x+y\|^2} \bigg) \ dx $$

for $x,y \in \mathbb R^d$. Does anyone know a solution or an approach I might take?

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  • $\begingroup$ Is it possible that the $(x+y)^2$ in the denominator should be $|x+y|^2$ instead? $\endgroup$ – Josiah Park Nov 18 '19 at 18:49
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    $\begingroup$ If so, write $|x+y|^2=2+2\langle x,y \rangle$ and just use the procedure from your previous question (find the first term in the Gegenbauer expansion of the function by integrating with respect to the measure which the Gegenbauer polynomials are orthogonal with respect to: mathworld.wolfram.com/GegenbauerPolynomial.html). $\endgroup$ – Josiah Park Nov 18 '19 at 18:54
  • $\begingroup$ Yes, sloppy notation. I'll fix the post. $\endgroup$ – user148767 Nov 18 '19 at 19:20
  • $\begingroup$ I see. Where $(1-t^2)^{\frac{d-2}{2}}$ is the measure? And perhaps a minus sign in the exponential? $\endgroup$ – user148767 Nov 18 '19 at 19:33
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    $\begingroup$ The usual expansion is $f(t)=\sum\limits_{n=0}^\infty \hat{f_n} \frac{n+\lambda}{\lambda}C_n(t)$, where $\lambda=\frac{d-1}{2}$. The coefficient $\hat{f}_0=\frac{\Gamma(\lambda+1)}{\Gamma(\lambda+\frac{1}{2})\Gamma(\frac{1}{2})}\int\limits_{-1}^1f(t)(1-t^2)^{\lambda-\frac{1}{2}}dt$. So that formula should account for any missing constant. This gives value $\frac{2(2-\frac{2}{e})}{\pi}$ for $d=3$. $\endgroup$ – Josiah Park Nov 18 '19 at 19:50

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