Appendix: Useful Integrals
In the paper we used several integrals whose expressions are given here. For completeness, the proofs are given in this appendix, although they are essentially boring technical computations. For d small enough, Maple or WolframAlpha can compute most of such integrals. However, for symbolic d, such software may handle only a couple of them. Proofs are below.
$$\begin{aligned}&\int _0^\infty e^{-c r^d} r^{2 d -1} \mathrm{d} r = \frac{1}{d \cdot c^2}. \end{aligned}$$
(9)
$$\begin{aligned}&\int _0^{\pi } \sin ^d \alpha \, \mathrm{d} \alpha = \frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)} \cdot \sqrt{\pi }. \end{aligned}$$
(10)
$$\begin{aligned}&\int _0^\pi \int _0^\pi \sin ^d\alpha _1 \sin ^d\alpha _2 |\cos \alpha _1 -\cos \alpha _2| \mathrm{d}\alpha _1 \mathrm{d}\alpha _2 = \frac{2^{2 d+4} {d!}^2}{(2 d + 2)!}. \end{aligned}$$
(11)
$$\begin{aligned}&\int _0^{2\pi } \int _0^{2\pi } \cos {\left( \alpha _1 - \alpha _2\right) } \mathrm{d} \alpha _2 \mathrm{d} \alpha _1 = 0. \end{aligned}$$
(12)
$$\begin{aligned}&\int _0^{\pi } \sin ^d \alpha \cos \alpha \mathrm{d} \alpha = 0 . \end{aligned}$$
(13)
$$\begin{aligned}&\int _0^\pi \int _0^\pi \sin ^d\alpha _1 \sin ^d\alpha _2 |\cos \alpha _1 -\cos \alpha _2| \cos \alpha _1 \cos \alpha _2 \mathrm{d}\alpha _1 \mathrm{d}\alpha _2 = -\frac{2^{2 d+4} {d!}^2}{(2 d + 3)!}.\nonumber \\ \end{aligned}$$
(14)
$$\begin{aligned}&\int _0^{2\pi } \int _0^{2\pi } \cos ^2(\alpha _1-\alpha _2) \mathrm{d}\alpha _2 \mathrm{d}\alpha _1 = 2\pi ^2. \end{aligned}$$
(15)
$$\begin{aligned}&\int _0^\pi \sin ^d (\alpha ) \cos ^2 (\alpha ) \mathrm{d}\alpha = \biggl (\frac{1}{d+2}\biggr ) \frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)}\cdot \sqrt{\pi }. \end{aligned}$$
(16)
$$\begin{aligned}&{\int _0^\pi \int _0^\pi \sin ^d\alpha _1 \sin ^d\alpha _2 \cos ^2\alpha _1 \cos ^2\alpha _2 |\cos \alpha _1 -\cos \alpha _2| \mathrm{d}\alpha _1 \mathrm{d}\alpha _2} \nonumber \\&\quad = \frac{2^{2 d+6} (7 d + 13) {(d+2)!}^2}{(d+1)^2 (d+2) (2 d + 6)!}. \end{aligned}$$
(17)
$$\begin{aligned}&\int _{a}^{b} \sin ^d \alpha \, \mathrm{d} \alpha \nonumber \\&\quad = {\left\{ \begin{array}{ll} \displaystyle {-\sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} } \frac{1}{i} \left( \frac{\Gamma ((d + 1)/2) \Gamma (1+i/2)}{\Gamma (1+d/2) \Gamma ((1+i)/2)}\right) \sin ^{i-1} (\alpha ) \cos (\alpha ) \Big |^{b}_{a}, &{} \hbox {if } d\hbox { is odd},\\ \displaystyle {-\sum _{\mathop {i \mathrm{\ even}}\limits ^{2\le i\le d}}} \frac{1}{i} \left( \frac{\Gamma ((d+1)/2) \Gamma (1+i/2)}{\Gamma (1+d/2) \Gamma ((1+i)/2)}\right) \sin ^{i-1} (\alpha ) \cos (\alpha ) \Big |^{b}_{a} + \frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)} {\frac{\alpha }{\sqrt{\pi }}}\Big |^{b}_{a} , &{} \hbox {if } d\hbox { is even}. \end{array}\right. }\nonumber \\ \end{aligned}$$
(18)
$$\begin{aligned} \int _0^{\pi } \alpha \sin ^d \alpha \cos \alpha \, \mathrm{d} \alpha = -\biggl (\frac{1}{d+1}\biggr ) \frac{\Gamma (d/2+1)}{\Gamma (1+(d+1)/2)} \cdot \sqrt{\pi }. \end{aligned}$$
(19)
Proofs
Proof of Equation (9)
Classic. \(\square \)
Proof of Equation (18)
Integration by parts gives
$$\begin{aligned} \int _{a}^{b} \sin ^d \alpha \, \mathrm{d} \alpha = -\cos (\alpha )\sin ^{d-1}(\alpha )\Big |^{b}_{a} + \int _{a}^{b} \cos ^2{\alpha } (d-1) \sin ^{d-2} (\alpha ) \, \mathrm{d} \alpha . \end{aligned}$$
Using the identity \(\cos ^2(\alpha ) = 1 - \sin ^2(\alpha )\) and simplifying gives
$$\begin{aligned} \int _{a}^{b} \sin ^d \alpha \, \mathrm{d} \alpha = -\Big (\frac{1}{d}\Big ) \cos (\alpha )\sin ^{d-1}(\alpha )\Big |^{b}_{a} + \Big (\frac{d-1}{d}\Big ) \int _{a}^{b}\sin ^{d-2} (\alpha ) \, \mathrm{d} \alpha . \end{aligned}$$
Now, using \(\int _{a}^{b} \sin ^0 \alpha \, \mathrm{d} \alpha = \alpha |^{b}_{a}\) and the identity
$$\begin{aligned} \prod _{k=1}^{k} {\frac{2 i-1}{2 i}} = \frac{\Gamma ({1}/{2}+k)}{\Gamma (k+1) \sqrt{\pi }}, \end{aligned}$$
by induction on d, we get that, for d even,
$$\begin{aligned} \int _{a}^{b} \sin ^d \alpha \, \mathrm{d} \alpha= & {} \displaystyle {-\sum _{\mathop {i \mathrm{\ even}}\limits ^{2\le i\le d}}} \frac{1}{i} \left( \frac{{\Gamma ((d+1)/2}/\big ({\Gamma (1+d/2)}\sqrt{\pi }\big ) }{ {\Gamma ((i+1)/2)}/\big (\Gamma (1+i/2)\sqrt{\pi } \big ) }\right) \sin ^{i-1} (\alpha ) \cos (\alpha )\Big |^{b}_{a} \nonumber \\&+ \frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)} {\frac{\alpha }{\sqrt{\pi }}}\Big |^{b}_{a} . \end{aligned}$$
Analogously, for d odd, we use \(\int _{a}^{b} \sin ^1 \alpha \, \mathrm{d} \alpha = -\cos (\alpha )\big |^{b}_{a} \) and the identity
$$\begin{aligned} \prod _{k=1}^{k} {\frac{2 i}{2 i+1}} = \frac{\Gamma (d+1) \sqrt{\pi }}{\Gamma (d+{3}/{2})}, \end{aligned}$$
to get by induction
$$\begin{aligned} \int _{a}^{b} \sin ^d \alpha \, \mathrm{d} \alpha =-\sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{1}{i} \biggl (\frac{(\Gamma ((d+1)/2)\sqrt{\pi })/\Gamma (1+d/2)}{(\Gamma ((i+1)/2)\sqrt{\pi })/\Gamma (1+i/2)}\biggr ) \sin ^{i-1} (\alpha ) \cos (\alpha )\bigr |^{b}_{a}. \end{aligned}$$
Grouping the two expressions above completes the proof. \(\square \)
Proof of Equation (10)
It is a direct corollary of (18). \(\square \)
Proof of Equation (19)
Integration by parts combined with (10) gives
$$\begin{aligned} \int _0^{\pi } \alpha \sin ^d \alpha \cos \alpha \, \mathrm{d} \alpha= & {} \frac{1}{d+1} \alpha \sin ^{d+1}(\alpha ) \Big |^{\pi }_0 - \frac{1}{d+1} \int _0^{\pi } \sin ^{d+1} \alpha \, \mathrm{d} \alpha \\= & {} - \frac{1}{d+1} \int _0^{\pi } \sin ^{d+1} \alpha \, \mathrm{d} \alpha \\= & {} -\Big ( \frac{1}{d+1} \Big ) \frac{\Gamma (d/2+1)}{\Gamma (1+(d+1)/2)} \cdot \sqrt{\pi }. \end{aligned}$$
\(\square \)
Proof of Equation (11)
First, we use the symmetry with respect to the line \(\alpha _1=\alpha _2\) to split the integral in two equal parts without absolute values.
$$\begin{aligned} \mathrm{Integral}= & {} 2 \int \int _{\mathop {\alpha _1 < \alpha _2}\limits ^{\alpha _1,\alpha _2\in [0,\pi ]^2}} \;\;\sin ^d\alpha _1 \sin ^d\alpha _2 (\cos \alpha _1 -\cos \alpha _2) \mathrm{d}\alpha _1 \mathrm{d}\alpha _2\\= & {} 2 \biggl ( \int _0^{\pi } \sin ^d \alpha _1 \cos \alpha _1 \int _{\alpha _1}^\pi \sin ^d \alpha _2 \mathrm{d}\alpha _2 \mathrm{d}\alpha _1 - \int _0^{\pi } \sin ^d \alpha _2 \cos \alpha _2 \int _0^{\alpha _2} \sin ^d\alpha _1 \mathrm{d}\alpha _1 \mathrm{d}\alpha _2 \biggr ) \\= & {} 2 \int _0^{\pi } \sin ^d x \cos x \biggl (\int _{x}^\pi \sin ^d y \mathrm{d}y- \int _0^x \sin ^d y \mathrm{d}y\biggr )\mathrm{d}x. \\ \end{aligned}$$
Let \(f_d(x) = (\int _{x}^\pi \sin ^d y \mathrm{d}y - \int _0^x \sin ^d y \mathrm{d}y)\). Then, by (18), we get
$$\begin{aligned} f_d(x) = {\left\{ \begin{array}{ll} \displaystyle {2 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} } \frac{1}{i} \left( \frac{\Gamma ((d+1)/2)\Gamma (1+i/2)}{\Gamma (1+d/2)\Gamma ((1 + i)/2)}\right) \sin ^{i-1} x \cos x, &{} \hbox {if } d\hbox { is odd},\\ \displaystyle {2 \sum _{\mathop {i \mathrm{\ even}}\limits ^{2\le i\le d}}} \frac{1}{i} \left( \frac{\Gamma ((d+1)/2)\Gamma (1+i/2)}{\Gamma (1+d/2)\Gamma ((1 + i)/2)}\right) \sin ^{i-1} x \cos x + \frac{\Gamma ((d + 1)/2)}{\Gamma (1+d/2)} \sqrt{\pi } {\Big (1 - \frac{2 x}{\pi }\Big )}, &{} \hbox {if } d\hbox { is even}. \end{array}\right. } \end{aligned}$$
First, suppose d is odd. Then, replacing the equation above in the simplified integral gives
$$\begin{aligned} 2 \int ^{\pi }_0 \sin ^{d} x \cos x \biggl ( 2 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{\Gamma ((d+1)/2) \Gamma (1+i/2)}{\Gamma (1+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \sin ^{i-1} x \cos x\biggr ) \mathrm{d} x, \end{aligned}$$
which simplifies into
$$\begin{aligned} 4 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{\Gamma ((d+1)/2)\Gamma (1+i/2)}{\Gamma (1+d/2)\Gamma ((i+1)/2)} \frac{1}{i} \int ^\pi _0 \sin ^{d+i-1} x \cos ^2 x \mathrm{d} x. \end{aligned}$$
Using (16) gives
$$\begin{aligned} 4 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{1}{i}\, \frac{\Gamma ((d+1)/2)\Gamma (1+i/2)}{\Gamma (1+d/2)\Gamma ((i+1)/2)} \biggl (\frac{1}{d+i+1}\biggr )\frac{\Gamma ((d+i)/2)}{\Gamma ((d+i+1)/2)} \,\sqrt{\pi }. \end{aligned}$$
Taking \(d = 2k-1\), \(i=2j-1\), replacing in the equation above and summing, gives
$$\begin{aligned} 4 \frac{(4k+ 1) \sqrt{\pi } \Gamma (k) \Gamma (2 k) }{\Gamma (2 k + {3}/{2}) \Gamma (1+k)}. \end{aligned}$$
Finally, replacing k by \((d+1)/2\) and simplifying gives
$$\begin{aligned} 2 \sqrt{\pi }\,\frac{\Gamma (d+1)}{(d+1)\Gamma (d+1+1/2)/2}. \end{aligned}$$
Now, suppose d is even. Then, again, replacing \(f_d\) in the simplified integral gives
$$\begin{aligned} 2 \int ^{\pi }_0 \sin ^{d} x \cos x&\Big ( \underbrace{2 \sum _{\mathop {i \mathrm{\ even}}\limits ^{2\le i\le d}} \frac{\Gamma ((d+1)/2)\Gamma (1+i/2)}{\Gamma (1+d/2)\Gamma ((i+1)/2) } \frac{1}{i} \sin ^{i-1} x \cos x}_{A} \\&+ \underbrace{\frac{\Gamma ((d+1)/2)}{\Gamma (1 + d/2)} \sqrt{\pi } {\Big (1 - \frac{2 x}{\pi }\Big )} }_{B}\Big ) \mathrm{d} x, \end{aligned}$$
We handle the A term first, which is pretty similar to the d odd case. Again, for the A term, expanding the sum and using (16) gives
$$\begin{aligned} 4 \sum _{\mathop {i \mathrm{\ even}}\limits ^{2\le i\le d}} \frac{1}{i}\, \frac{\Gamma ((d+1)/2)\Gamma (1+i/2)}{\Gamma (1+d/2)\Gamma ((i+1)/2)}\biggl (\frac{1}{d+i+1}\biggr ) \frac{\Gamma ((d+i)/2)}{\Gamma ((d+i+1)/2)}\, \sqrt{\pi }. \end{aligned}$$
Taking \(d = 2k\), \(i=2j\), replacing in the equation above and summing, gives
$$\begin{aligned} 4 \Big ( \frac{\sqrt{\pi } \Gamma (2 k + 1)}{(2 k + 1) \Gamma (2k + {3}/{2})} - \frac{2}{(2 k + 1)^2} \Big ). \end{aligned}$$
Finally, replacing k by d / 2 and simplifying gives
$$\begin{aligned} 2 \sqrt{\pi }\, \frac{\Gamma (d+1)}{(d+1)\Gamma (d+1+1/2)/2} - 2 \,\frac{1}{(d+1)^2/4}. \end{aligned}$$
Now, we handle the B term, which is
$$\begin{aligned} 2 \int ^\pi _0 \sin ^d x \cos x \Big (\frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)}\Big ) \sqrt{\pi } \Big (1 - \frac{2 x}{\pi } \Big ) \mathrm{d} x. \end{aligned}$$
Expanding it, gives
$$\begin{aligned} 2 \int ^\pi _0 \sin ^d x \cos x \Big (\frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)}\Big ) \sqrt{\pi } \mathrm{d} x - 2 \cdot \frac{2}{\pi } \cdot \int ^\pi _0 x \sin ^d x \cos x \Big (\frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)}\Big ) \sqrt{\pi } \mathrm{d} x. \end{aligned}$$
The left term is null because of (13). And plugging (19) in the right term above gives
$$\begin{aligned} 2 \cdot \frac{\Gamma ((d+1)/2)}{\Gamma (1+d/2)} \sqrt{\pi } \cdot \Big (\frac{1}{d+1}\Big ) \cdot \frac{\Gamma (1+d/2)}{\Gamma (1+(d+1)/2)} \cdot \sqrt{\pi } \cdot \frac{2}{\pi } = 2 \,\frac{1}{(d+1)^2/4}. \end{aligned}$$
Summing A term with B term gives
$$\begin{aligned} 2 \sqrt{\pi }\, \frac{\Gamma (d+1)}{(d+1) \Gamma (d+1+1/2)/2}. \end{aligned}$$
Therefore, the expression above holds for both odd and even d. Moreover, it can be simplified to the following nicer expression
$$\begin{aligned} \frac{2^{2d + 4} (d!)^2}{(2d+2)!}. \end{aligned}$$
\(\square \)
Proof of Equation (12)
Symmetry on the circle gives 0. \(\square \)
Proof of Equation (13)
Symmetry with respect to \(\pi /2\) gives 0. \(\square \)
Proof of Equation (14)
As in the proof of (11), we simplify this integral to get rid of the absolute value and isolate each of its variable.
$$\begin{aligned} \mathrm{Integral}= & {} 2 \int \int _{\mathop {\alpha _1 < \alpha _2}\limits ^{\alpha _1,\alpha _2\in [0,\pi ]^2}} \;\;\sin ^d\alpha _1 \sin ^d\alpha _2 (\cos \alpha _1 -\cos \alpha _2) \cos \alpha _1\cos \alpha _2 \mathrm{d}\alpha _1 \mathrm{d}\alpha _2\\= & {} 2 \Big ( - \int _0^{\pi } \sin ^d \alpha _1 \cos \alpha _1 \int _{\alpha _1}^\pi \sin ^d \alpha _2 \cos ^2 \alpha _2\mathrm{d}\alpha _2 \mathrm{d}\alpha _1 \\&\quad + \int _0^{\pi } \sin ^d \alpha _2 \cos \alpha _2 \int _0^{\alpha _2} \sin ^d \alpha _1 \cos ^2 \alpha _1 \mathrm{d}\alpha _1 \mathrm{d}\alpha _2\Big ) \\= & {} 2 \int _0^{\pi } \sin ^d x \cos x \Big (-\int _{x}^\pi \sin ^d y\cos ^2 y \mathrm{d}y + \int _0^x \sin ^d y\cos ^2 y \mathrm{d}y\Big ). \\= & {} -2 \int _0^{\pi } \sin ^d x \cos x f_d(x) \mathrm{d}x + 2 \int _0^{\pi } \sin ^d x \cos x f_{d+2}(x) \mathrm{d}x , \end{aligned}$$
where \(f_d\) is defined in the proof of (11). The first integral is exactly the opposite as the one in (11), the second integral is quite similar, for d odd it gives:
$$\begin{aligned}&2 \int ^{\pi }_0 \sin ^{d} x \cos x \Big ( 2 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d+2}} \frac{\Gamma ((d+3)/2) \Gamma (1+i/2)}{\Gamma (2+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \sin ^{i-1} x \cos x\Big ) \mathrm{d} x \\&\quad = 4 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d+2}} \frac{\Gamma ((d+3)/2)\Gamma (1+i/2)}{\Gamma (2+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \int ^{\pi }_0 \sin ^{d+i-1} x \cos ^2 x\;\mathrm{d}x\;\\&\quad = 4 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d+2}} \frac{\Gamma ((d+3)/2)\Gamma (1+i/2)}{\Gamma (2+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \big (\frac{1}{d+i+1} \big ) \frac{\Gamma ((d+i)/2)}{\Gamma \big ((d + i + 1)/2 \big )} \sqrt{\pi }\\&\quad = \frac{\Gamma (d-1)}{\Gamma (d+1/2)}. \end{aligned}$$
Substracting the two integrals yields the claimed result. The case with d even can be solved similarly. \(\square \)
Proof of Equation (15)
Simple integration. \(\square \)
Proof of Equation (16)
Use \(\cos ^2 x = 1-\sin ^2 x\) and (10), then simplify. \(\square \)
Proof of Equation (17)
As in the proof of (11) and (14), we simplify this integral to get rid of the absolute value and isolate each of its variable.
$$\begin{aligned} \mathrm{Integral}= & {} 2 \int \int _{\mathop {\alpha _1 < \alpha _2}\limits ^{\alpha _1,\alpha _2\in [0,\pi ]^2}} \;\;\sin ^d\alpha _1 \sin ^d\alpha _2 (\cos \alpha _1 -\cos \alpha _2) \cos ^2\alpha _1\cos ^2\alpha _2 \mathrm{d}\alpha _1 \mathrm{d}\alpha _2\\= & {} 2 \Big ( \int _0^{\pi } \sin ^d \alpha _1 \cos ^3 \alpha _1 \int _{\alpha _1}^\pi \sin ^d \alpha _2 \cos ^2 \alpha _2\mathrm{d}\alpha _2 \mathrm{d}\alpha _1 \\&- \int _0^{\pi } \sin ^d \alpha _2 \cos ^3 \alpha _2 \int _0^{\alpha _2} \sin ^d \alpha _1 \cos ^2 \alpha _1 \mathrm{d}\alpha _1 \mathrm{d}\alpha _2 \Big ) \\= & {} 2 \int _0^{\pi } \sin ^d x \cos x(1-\sin ^2 x) \Big (\int _{x}^\pi \sin ^d y(1-\sin ^2 y) \mathrm{d}y \\&- \int _0^x \sin ^d y(1-\sin ^2 y) \mathrm{d}y\Big ). \\= & {} 2 \int _0^{\pi } \sin ^d x \cos x f_d(x) \mathrm{d}x - 2 \int _0^{\pi } \sin ^d x \cos x f_{d+2}(x) \mathrm{d}x\\&\qquad \qquad - 2 \int _0^{\pi } \sin ^{d+2} x \cos x f_d(x) \mathrm{d}x + 2 \int _0^{\pi } \sin ^{d+2} x \cos x f_{d+2}(x) \mathrm{d}x , \end{aligned}$$
where \(f_d\) is defined in the proof of (11). The first and fourth integrals are the same as the ones in (11) (with d replaced by \(d+2\) in the fourth integral), the second integral is the same as the one in (14), the third integral is quite similar, for d odd the third integral gives:
$$\begin{aligned}&2 \int ^{\pi }_0 \sin ^{d+2} x \cos x \biggl ( 2 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{\Gamma ((d+1)/2) \Gamma (1+i/2)}{\Gamma (1+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \sin ^{i-1} x \cos x\biggr ) \mathrm{d} x \\&\quad = 4 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{\Gamma ((d+1)/2) \Gamma (1+i/2)}{\Gamma (1+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \int ^{\pi }_0 \sin ^{d+i+1} x \cos ^2 x\\&\quad = 4 \sum _{\mathop {i \mathrm{\ odd}}\limits ^{1\le i\le d}} \frac{\Gamma ((d+1)/2) \Gamma (1+i/2)}{\Gamma (1+d/2) \Gamma ((i+1)/2)} \frac{1}{i} \Big (\frac{1}{d+i+3} \Big ) \frac{\Gamma ((d+i+2)/2)}{\Gamma ((d+i+3)/2)}\, \sqrt{\pi }\\&\quad = \frac{4\;\Gamma (d+3)}{(d+3)\Gamma (d+7/2)}. \end{aligned}$$
Adding the four integrals yields the claimed result. The case with d even can be solved similarly. \(\square \)
