On Some Integrals Over A Unit Sphere Tau¯quar R. Khan1 Clemson University Clemson, SC 29634-0975 [email protected] Mark A. Pinsky2 Northwestern University Evanston, IL 60208-2730 [email protected] n n¡1 n Let Bn ½ R be a unit ball and S ½ R be a unit sphere. For n = 2, the unit ball is the interior of the unit circle mainly B2 = f(x1; x2) 2 2 2 2 2 R : x1 + x2 · 1g (see Figure 1a). The unit sphere in R is the unit circle 1 2 2 2 S = f(x1; x2) 2 R : x1 + x2 = 1g (see Figure 1b). Similarly for n = 3, the 3 2 2 2 unit ball is given by B3 = f(x1; x2; x3) 2 R : x1 + x2 + x3 · 1g and the unit 2 3 2 2 2 sphere is the spherical shell S = f(x1; x2; x3) 2 R : x1 + x2 + x3 = 1g (see Figure 2ab). Of course, in higher dimension, it is di±cult to visualize these geometric objects however the mathematical de¯nition still holds true. One can still de¯ne a point on the unit sphere as ! = (x1; x2; : : : ; xn) as a unit n¡1 2 Pn 2 vector on S with j!j = j=1 xj = 1. Integrals over the unit sphere in Rn are common in solving problems in spherical coordinates where azimuthal symmetry is not present. For exam- m ple, spherical harmonics in n = 3 are usually denoted by Yl (θ; Á) where normally θ is taken as the polar (colatitudinal) coordinate with θ 2 [0; ¼], and Á as the azimuthal (longitudinal) coordinate with Á 2 [0; 2¼), see Ar- m fken [1] or Pinsky2 [9]. In 3-D The spherical harmonics Yl is usually the angular portion of the solution for a boundary value problems involving a 1T. Khan is an Assistant Professor in the Department of Mathematical Sciences at Clemson University. 2M. Pinsky is a Professor in the Department of Mathematics at Northwestern Univer- sity. 1 1 1 0.5 0.5 ±1 ±0.5 0.5 1 ±1 ±0.5 0.5 1 ±0.5 ±0.5 ±1 ±1 (a) (b) Figure 1: (a) x2 + y2 · 1 (b) x2 + y2 = 1 (Using Maple) spherical (a) (b) Figure 2: (a) x2 + y2 + z2 · 1 (b) x2 + y2 + z2 = 1 (Using Maple) 2 partial di®erential equation in spherical coordinates. Spherical harmonics are implemented in Mathematica as SphericalHarmonicY[l, m, theta, phi]. The normalized spherical harmonics actually form an orthonormal basis so any arbitrary real function f(θ; Á) can be expanded in terms of the spher- ical harmonics and one can ¯nd Fourier series type expansions using these functions. The calculation of the coe±cient of the expansion involves inte- grating the function over the unit sphere in S2. The applications of spherical harmonics range from solving Laplace's equation in physics and engineering to solving radiative transport equations for biological tissues. See Gri±ths [2] for some examples in electro statics involving Laplace's equations and see Ishimaru [3] for some examples in scattering of photons in random media involving radiative transport equations. The motivation for our note comes from using spherical harmonics for solving the photon transport equation in biological tissue with spatially varying refractive index, see Ferwerda [4] or Khan [5]. The radiative transport equation (RTE) for a medium with a spatially varying refractive index describes the density of particles (photons) which travel at time t though the point y in the direction ! 2 Sn¡1. One can derive simpler deterministic models from the RTE by expanding the density in spherical harmonics and retaining a limited number of terms. This ex- pansion also requires integrating the radiative transport equation over Sn¡1. Now that we have stated our motivation, the rest of the paper will deal with integrals over the unit sphere. Common Integrals Over A Unit Sphere. There are several common integrals over a unit sphere in Rn. The simplest is the integral of a constant over a unit sphere: Z 2(¼)n=2 d! = jSn¡1j = (1) Sn¡1 ¡(n=2) where d! is the surface measure on Sn¡1 and Z 1 ¡(m) = ym¡1e¡ydy 0 3 is the Gamma function which satis¯es ¡(m + 1) = m¡(m) if m > 0 and p ¡(m + 1) = m! if m is an integer with ¡(2) = ¡(1) = 1 and ¡(1=2) = ¼. For n = 2, d! is given as the incremental length dθ on the unit circle and if we integrate around the unit circle we get jS1j = 2¼ which is the circumference of the circle. For n = 3, d! is the incremental area dθdÁ on the spherical shell and if we integrate around the spherical shell we get jS2j = 4¼ which is the area of the spherical shell. These values of course corresponds to equation (1) as one can easily see from the properties of the ¡ function just mentioned. The next two integrals are slightly more complicated. R n¡1 jS j n Proposition 1 Sn¡1 (a ¢ !)!d! = n a for any vector a in R . T n¡1 Proof: Let ! = (x1; x2; : : : ; xn) 2 S so that j!j = 1 and x1 6= ¡1, then one can construct an orthogonal basis f!; !2;:::;!ng ½ Sn¡1 in Rn n (see Cullen [6]). Let fe1; e2; : : : ; eng be the standard basis in R and denote G 2 SO(n) (special orthogonal group) which permutes the standard basis as 2 n follows:Ge1 = e2; Ge2 = e3; : : : ; Gen = e1. Then if we let X = (!; ! ;:::;! ) and H = XGX¡1. Then obviously H 2 SO(n) and H! = !2;H!2 = !3;:::;H!n = !. Therefore f!; H!; H2!; : : : ; Hn¡1!g forms an orthogonal n Pn¡1 k k basis in R . Hence, any vector a can be written as a = k=0(a ¢ H !)H ! and Z Z Z 1 Xn¡1 1 jSn¡1j (a ¢ !)!d! = (a ¢ Hk!)Hk!dHk! = ad! = a: (2) n¡1 n n¡1 n n¡1 n S k=0 S S R jSn¡1j Proposition 2 Sn¡1 (a ¢ !)(b ¢ !)d! = n (a ¢ b) for any vectors a and b in Rn. Proof: Let f!; H!; : : : ; Hn¡1!g ½ Sn¡1 be a basis in Rn as in the last proof. Pn¡1 k k Hence any two vectors a and b can be written as a = k=0(a¢H !)H ! and Pn¡1 k k Pn¡1 k k b = k=0(b¢H !)H !. If we further note that a¢b = k=0(a¢H !)(b¢H !), we get Z Z 1 Xn¡1 (a ¢ !)(b ¢ !)d! = (a ¢ Hk!)(b ¢ Hk!)dHk! n¡1 n n¡1 S k=0 S 4 Z 1 jSn¡1j = (a ¢ b)d! = (a ¢ b): (3) n Sn¡1 n Uncommon Integrals Over A Unit Sphere. The integrals explained below arise in ¯nding the spherical harmonic expansions for the radiative transport equation with spatially varying refractive indices, see Ferwerda [4] ¡1 ¡1 ¡1 ¡1 and Khan [5]. Let ! = (x1 ; x2 ; : : : ; xn ), we will now investigate how to evaluate integrals of this form, Z (a ¢ !¡1)!d!: (4) Sn¡1 First, we note that these integrals are only de¯ned in the Cauchy principal value sense. Let us ¯rst discuss the simplest of the cases n = 2; 3 to clarify the meaning of these integrals. R ¡1 1 2 Proposition 3 S1 (a ¢ ! )!d! = jS ja for any vector a in R . ¡1 1 1 Proof: ! = cos(Á) x^ + sin(Á) y^, wherex ^,^y are unit vectors in the xy plane. Therefore, µ ¶ µ ¶ ¡1 x1 x2 (a ¢ ! )! = ax + ay x^ + ax + ay y^ (5) x2 x1 x1 x2 where x1 = cos(Á) and x2 = sin(Á). Now if we compute and and x2 x1 simplify the right hand side of equation (4) we get, ¡1 (a ¢ ! )! = a + ay cot(Á)^x + ax tan(Á)^y: (6) Therefore, Z Z Z Z ¡1 (a ¢ ! )!d! = ad! + ay cot(Á)d!x^ + ax tan(Á)d!y:^ (7) 2¼ 2¼ 2¼ 2¼ Since Cauchy's principal value integrals (see Brown and Churchill [7]) of cot(Á) and tan(Á) over all of 2¼ radians vanish, we get, Z (a ¢ !¡1)!d! = 2¼a: (8) 2¼ 5 R ¡1 2 3 Proposition 4 S2 (a ¢ ! )!d! = jS ja for any vector a in R . ¡1 1 1 1 Proof: ! = sin(θ) cos(Á) x^ + sin(θ) sin(Á) y^ + cos(θ) z^, wherex ^,^y; z^ are unit vectors in the xyz space. Therefore, µ ¶ µ ¶ x x x x (a ¢ !¡1)! = a + 1 a + 1 a x^ + 2 a + a + 2 a y^ x x y x z x x y x z 2 µ3 1 ¶ 3 x3 x3 + ax + ay + az z^ (9) x1 x2 where x1 = sin(θ) cos(Á), x2 = sin(θ) sin(Á), and x3 = cos(θ). Now if we compute x1 , x1 , x2 , x2 , x3 , and x3 and simplify the right hand side of equation x2 x3 x1 x3 x1 x2 (9) we get, ¡1 (a ¢ ! )! = a + (ay cot(Á) + az tan(θ) cos(Á))^x + (ax tan(Á) + az tan(θ) sin(Á))^y + (ax cot(θ) sec(Á) + ay cot(θ) csc(Á))^z: Therefore, Z Z Z ¡1 (a ¢ ! )!d! = ad! + (ay cot(Á) + az tan(θ) cos(Á))d!x^ 4¼ Z4¼ 4¼ + (ax tan(Á) + az tan(θ) sin(Á))d!y^ Z4¼ + (ax cot(θ) sec(Á) + ay cot(θ) csc(Á))d!z^ (10) 4¼ Since Cauchy's principal value integrals of cot(Á), tan(θ) cos(Á),tan(Á),tan(θ) sin(Á), cot(θ) sec(Á), and cot(θ) csc(Á) over all of 4¼ steradians vanish, we get, Z (a ¢ !¡1)!d! = 4¼a: (11) 4¼ R ¡1 1 Proposition 5 S1 (a ¢ ! )(b ¢ !)d! = jS j(a ¢ b) for any vectors a and b in R2.