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UPDATE

I have a conjecture as to the solution, based on some theoretical considerations, and it holds at least for $d=3,4$: $$ \int_0^\pi \left( \frac{\sin\phi}{\sqrt{1+u^2-2u\cos\phi}}\right)^{d-2}\,d\phi = \frac{\omega_{d}}{2\omega_{d-1}} \frac{u^{d-2}+1 - |u^{d-2}-1|}{u^{d-2}} = \frac{\omega_d}{\omega_{d-1}}\begin{cases} 1 & u\le1 \\ u^{2-d} & u>1 \end{cases} $$ Notice that $\int_0^\pi \sin^{d-2}\phi \,d\phi = \frac{\omega_d}{\omega_{d-1}}$. Not sure how to prove the conjecture though.

Background

I'm interested in computing a certain $d$-dimensional integral, for $d\ge3$, over the unit ball $B_d(\mathbf0,1)$. Let's denote its surface area by $\omega_d$; we know that its volume is then $\omega_d / d$. Then, the integral is: $$H(x) := \frac{d}{(d-2)\omega_d^2} \int_{B_d(\mathbf0,1)} |x-y|^{2-d}\,dy~~.$$ Using the usual spherical coordinates, let's write $x = (r,\theta)$ and $y = (\rho,\phi)$, the angles measured from the $d^\text{th}$ coordinate. Then we have \begin{align*} H(r,\theta) &= \frac{d}{(d-2)\omega_d^2}\int_0^1\int_{S_d(\mathbf0,1)} \left(r^2+\rho^2 - 2r\rho\cos(\theta-\phi) \right)^{\frac{2-d}{2}}\,d\Omega\,\rho^{d-1}d\rho\\ &= \frac{dr^{2-d}}{(d-2)\omega_d^2}\int_0^1\int_{S_d(\mathbf0,1)} \left(1+(\rho/r)^2 - 2(\rho/r)\cos(\theta-\phi) \right)^{\frac{2-d}{2}}\,d\Omega\,\rho^{d-1}d\rho\\ &= \frac{dr^2}{(d-2)\omega_d^2}\int_0^{1/r}\int_{S_d(\mathbf0,1)}\left( 1+u^2 - 2u\cos\phi\right)^{\frac{2-d}{2}}\,d\Omega\,u^{d-1}du\\ &= \frac{dr^2\omega_{d-1}}{(d-2)\omega_d^2}\int_0^{1/r}\int_0^\pi \left( \frac{\sin\phi}{\sqrt{1+u^2-2u\cos\phi}}\right)^{d-2}\,d\phi\,u^{d-1}du~~. \end{align*}

Question

For general integer $d\ge3$, how does one solve the following integral for positive $u$: $$ \int_0^\pi \left( \frac{\sin\phi}{\sqrt{1+u^2-2u\cos\phi}}\right)^{d-2}\,d\phi~~. $$ For $d=3$, the substitution $t = -\cos\phi$ solves the problem succinctly, but for any greater $d$, it looks like some complex techniques may be necessary (e.g., as in $\int_0^{2\pi}\log(1 + u^2 - 2u\cos\phi)\,d\phi$, which is the analogous expression for $d=2$). However, my attempts in this direction haven't borne fruit. Any help is appreciated!

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The dependency on $u$ in your conjecture appears incorrect. I derive the case for odd dimension $d$ below. (The even case yields an integral of a rational function, and hence is easier to evaluate.)

$$\begin{align} G{\left(d;u\right)} &=\int_{0}^{\pi}\left(\frac{\sin{\phi}}{\sqrt{1+u^2-2u\cos{\phi}}}\right)^{d-2}\,\mathrm{d}\phi\\ &=\int_{0}^{\pi}\frac{\sin^{d-2}{\phi}}{\left(1+u^2-2u\cos{\phi}\right)^{\frac{d}{2}-1}}\,\mathrm{d}\phi\\ G{\left(d;\tan{\left(\frac{t}{2}\right)}\right)}&=\int_{0}^{\pi}\frac{\sin^{d-2}{\phi}}{\left[\sec^2{\left(\frac{t}{2}\right)}\left(1-\sin{t}\cos{\phi}\right)\right]^{\frac{d}{2}-1}}\,\mathrm{d}\phi\\ &=\cos^{d-2}{\left(\frac{t}{2}\right)}\int_{0}^{\pi}\frac{\sin^{d-2}{\phi}}{\left(1-\sin{t}\cos{\phi}\right)^{\frac{d}{2}-1}}\,\mathrm{d}\phi\\ \end{align}$$

Case: $d$ odd

$$\begin{align} G{\left(2m+3;\tan{\left(\frac{t}{2}\right)}\right)} &=\cos^{2m+1}{\left(\frac{t}{2}\right)}\int_{0}^{\pi}\frac{\sin^{2m+1}{\phi}}{\left(1-\sin{t}\cos{\phi}\right)^{m+\frac12}}\,\mathrm{d}\phi\\ &=\cos^{2m+1}{\left(\frac{t}{2}\right)}\int_{-1}^{1}\frac{\left(1-v^2\right)^{m}}{\left(1-v\sin{t}\right)^{m+\frac12}}\,\mathrm{d}v\\ &=\cos^{2m+1}{\left(\frac{t}{2}\right)}\int_{0}^{1}\frac{\left[4w(1-w)\right]^{m}}{\left(1+\sin{t}-2w\sin{t}\right)^{m+\frac12}}\,2\,\mathrm{d}w;~~[v=2w-1]\\ &=\frac{2^{2m+1}\cos^{2m+1}{\left(\frac{t}{2}\right)}}{\left(1+\sin{t}\right)^{m+\frac12}}\int_{0}^{1}\frac{w^m(1-w)^{m}}{\left(1-2w\frac{\sin{t}}{1+\sin{t}}\right)^{m+\frac12}}\,\mathrm{d}w\\ &=\frac{2^{2m+1}\cos^{2m+1}{\left(\frac{t}{2}\right)}}{\left(1+\sin{t}\right)^{m+\frac12}}\int_{0}^{1}\frac{w^m(1-w)^{m}}{\left(1-zw\right)^{m+\frac12}}\,\mathrm{d}w\\ &=\frac{2^{2m+1}\cos^{2m+1}{\left(\frac{t}{2}\right)}}{\left(1+\sin{t}\right)^{m+\frac12}}\frac{\Gamma{\left(m+1\right)}^2}{\Gamma{\left(2m+2\right)}}\,{_2F_1}{\left(m+\frac12,m+1;m+2;z\right)}\\ &=\frac{2^{2m+1}\cos^{2m+1}{\left(\frac{t}{2}\right)}}{\left(1+\sin{t}\right)^{m+\frac12}}\frac{\Gamma{\left(m+1\right)}^2}{\Gamma{\left(2m+2\right)}}\,\frac{2^{2m+1}}{\left(1+\sqrt{1-z}\right)^{2m+1}}\\ G{\left(2m+3;u\right)}&=\frac{2^{2m+1}}{(u+1)^{2m+1}}\frac{\Gamma{\left(m+1\right)}^2}{\Gamma{\left(2m+2\right)}}\,\frac{2^{2m+1}}{\left(1+\frac{\left|u-1\right|}{u+1}\right)^{2m+1}}\\ G{\left(2m+3;u\right)}&=\frac{\Gamma{\left(m+1\right)}^2}{\Gamma{\left(2m+2\right)}}\,\frac{4^{2m+1}}{\left(u+1+\left|u-1\right|\right)^{2m+1}}\\ &=\frac{\sqrt{\pi}\,\Gamma{\left(m+1\right)}}{\Gamma{\left(m+\frac32\right)}}\,\frac{2^{2m+1}}{\left(u+1+\left|u-1\right|\right)^{2m+1}}\\ G{\left(d,u\right)}&=\frac{\sqrt{\pi}\,\Gamma{\left(\frac{d-1}{2}\right)}}{\Gamma{\left(\frac{d}{2}\right)}}\,\frac{2^{d-2}}{\left(u+1+\left|u-1\right|\right)^{d-2}}.\\ \end{align}$$

As you can see, this matches your conjecture for $d-3$, but in general this won't be the case.

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  • $\begingroup$ Actually, the functional form of the conjecture was wrong, but the desired behavior (flat for $u<1$, then decaying like $u^{2-d}$ for $u>1$) matches perfectly, even up to the constant! Thank you very much! $\endgroup$ – Eugene Shvarts Feb 11 '15 at 20:12

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