Appendix A Gamma and Beta Functions A.1 A Useful Formula The following formula is valid: Z 2 √ n e−|x| dx = π . Rn This is an immediate consequence of the corresponding one-dimensional identity Z +∞ 2 √ e−x dx = π , −∞ which is usually proved from its two-dimensional version by switching to polar coordinates: Z +∞ Z +∞ 2 2 Z ∞ 2 I2 = e−x e−y dydx = 2π re−r dr = π . −∞ −∞ 0 A.2 Definitions of Γ (z) and B(z,w) For a complex number z with Rez > 0 define Z ∞ Γ (z) = tz−1e−t dt. 0 Γ (z) is called the gamma function. It follows from its definition that Γ (z) is analytic on the right half-plane Rez > 0. Two fundamental properties of the gamma function are that Γ (z + 1) = zΓ (z) and Γ (n) = (n − 1)!, where z is a complex number with positive real part and n ∈ Z+. Indeed, integration by parts yields Z ∞ tze−t ∞ 1 Z ∞ 1 Γ (z) = tz−1e−t dt = + tze−t dt = Γ (z + 1). 0 z 0 z 0 z Since Γ (1) = 1, the property Γ (n) = (n − 1)! for n ∈ Z+ follows by induction. Another important fact is that 417 418 A Gamma and Beta Functions √ 1 Γ 2 = π . This follows easily from the identity Z ∞ 1 Z ∞ 2 √ 1 − 2 −t −u Γ 2 = t e dt = 2 e du = π . 0 0 Next we define the beta function. Fix z and w complex numbers with positive real parts. We define Z 1 Z 1 B(z,w) = tz−1(1 −t)w−1 dt = tw−1(1 −t)z−1 dt. 0 0 We have the following relationship between the gamma and the beta functions: Γ (z)Γ (w) B(z,w) = , Γ (z + w) when z and w have positive real parts. The proof of this fact is as follows: Z 1 Γ (z + w)B(z,w) = Γ (z + w) tw−1(1 −t)z−1 dt 0 Z ∞ 1 z+w = Γ (z + w) uw−1 du t = u/(1 + u) 0 1 + u Z ∞ Z ∞ 1 z+w = uw−1 vz+w−1e−v dvdu 0 0 1 + u Z ∞ Z ∞ = uw−1sz+w−1e−s(u+1) dsdu s = v/(1 + u) 0 0 Z ∞ Z ∞ = sze−s (us)w−1e−su duds 0 0 Z ∞ = sz−1e−sΓ (w)ds 0 = Γ (z)Γ (w). A.3 Volume of the Unit Ball and Surface of the Unit Sphere n We denote by vn the volume of the unit ball in R and by ωn−1 the surface area of the unit sphere Sn−1. We have the following: n 2π 2 ωn−1 = n Γ ( 2 ) and A.4 Computation of Integrals Using Gamma Functions 419 n n ωn−1 2π 2 π 2 vn = = n = n . n nΓ ( 2 ) Γ ( 2 + 1) The easy proofs are based on the formula in Appendix A.1. We have Z Z ∞ √ n −|x|2 −r2 n−1 π = e dx = ωn−1 e r dr , Rn 0 by switching to polar coordinates. Now change variables t = r2 to obtain that Z ∞ n ω n ω 2 n−1 −t 2 −1 n−1 n π = 2 e t dt = 2 Γ 2 . 0 This proves the formula for the surface area of the unit sphere in Rn. To compute vn, write again using polar coordinates Z Z Z 1 n−1 1 vn = |B(0,1)| = 1dx = r dr dθ = ωn−1 . |x|≤1 Sn−1 0 n Here is another way to relate the volume to the surface area. Let B(0,R) be the ball in Rn of radius R > 0 centered at the origin. Then the volume of the shell B(0,R + h) \ B(0,R) divided by h tends to the surface area of B(0,R) as h → 0. In other words, the derivative of the volume of B(0,R) with respect to the radius R is n equal to the surface area of B(0,R). Since the volume of B(0,R) is vnR , it follows n−1 that the surface area of B(0,R) is nvnR . Taking R = 1, we deduce ωn−1 = nvn. A.4 Computation of Integrals Using Gamma Functions Let k1,...,kn be nonnegative even integers. The integral n n Z 2 Z +∞ 2 k + k1 kn −|x| k j −x j 1 x ···x e dx1 ···dxn = x e j dx j = Γ n 1 n ∏ j ∏ R j=1 −∞ j=1 2 expressed in polar coordinates is equal to Z Z ∞ 2 k1 kn k1+···+kn n−1 −r θ1 ···θn dθ r r e dr , Sn−1 0 where θ = (θ1,...,θn). This leads to the identity Z −1 n k + k1 kn k1 + ··· + kn + n j 1 θ ···θn dθ = 2Γ Γ . n−1 1 ∏ S 2 j=1 2 Another classical integral that can be computed using gamma functions is the following: 420 A Gamma and Beta Functions Z π/2 1 Γ ( a+1 )Γ ( b+1 ) (sinϕ)a(cosϕ)b dϕ = 2 2 , 2 a+b+2 0 Γ ( 2 ) whenever a and b are complex numbers with Rea > −1 and Reb > −1. Indeed, change variables u = (sinϕ)2; then du = 2(sinϕ)(cosϕ)dϕ, and the pre- ceding integral becomes a+1 b+1 1 Z 1 a−1 b−1 1 a + 1 b + 1 1 Γ ( )Γ ( ) u 2 (1 − u) 2 du = B , = 2 2 . 2 2 2 2 2 a+b+2 0 Γ ( 2 ) A.5 Meromorphic Extensions of B(z,w) and Γ (z) Using the identity Γ (z + 1) = zΓ (z), we can easily define a meromorphic exten- sion of the gamma function on the whole complex plane starting from its known values on the right half-plane. We give an explicit description of the meromorphic extension of Γ (z) on the whole plane. First write Z 1 Z ∞ Γ (z) = tz−1e−t dt + tz−1e−t dt 0 1 and observe that the second integral is an analytic function of z for all z ∈ C. Write the first integral as ( ) Z 1 N (−t) j N (−1) j/ j! tz−1 e−t − ∑ dt + ∑ . 0 j=0 j! j=0 z + j The last integral converges when Rez > −N − 1, since the expression inside the curly brackets is O(tN+1) as t → 0. It follows that the gamma function can be de- fined to be an analytic function on Rez > −N − 1 except at the points z = − j, (−1) j j = 0,1,...,N, at which it has simple poles with residues j! . Since N was arbi- trary, it follows that the gamma function has a meromorphic extension on the whole plane. In view of the identity Γ (z)Γ (w) B(z,w) = , Γ (z + w) the definition of B(z,w) can be extended to C × C. It follows that B(z,w) is a mero- morphic function in each argument. A.6 Asymptotics of Γ (x) as x → ∞ We now derive Stirling’s formula: A.7 Euler’s Limit Formula for the Gamma Function 421 Γ (x + 1) lim x√ = 1. x→∞ x e 2πx q 2 First change variables t = x + sx x to obtain q x x 2 Z ∞ x √ Z +∞ 1 + s x Γ (x + 1) = e−ttx dt = 2x √ √ ds. 0 e − x/2 e2s x/2 p x Setting y = 2 , we obtain y 2y +∞ s ! Γ (x + 1) Z 1 + y √ = χ (s)ds. x x es (−y,∞) e 2x −∞ √ To show that the last integral converges to π as y → ∞, we need the following: (1) The fact that y 2y 1 + s/y 2 lim → e−s , y→∞ es which follows easily by taking logarithms and applying L’Hopital’sˆ rule twice. (2) The estimate, valid for y ≥ 1, (1 + s)2 s y !2y when s ≥ 0, 1 + es y ≤ es 2 e−s when −y < s < 0, which can be easily checked using calculus. Using these facts, the Lebesgue dom- inated convergence theorem, the trivial fact that χ−y<s<∞ → 1 as y → ∞, and the identity in Appendix A.1, we obtain that y 2y +∞ s ! Γ (x + 1) Z 1 + y lim x√ = lim χ(−y,∞)(s)ds x→∞ x y→∞ es e 2x −∞ Z +∞ 2 = e−s ds √−∞ = π. A.7 Euler’s Limit Formula for the Gamma Function For n a positive integer and Rez > 0 we consider the functions 422 A Gamma and Beta Functions Z n n t z−1 Γn(z) = 1 − t dt 0 n We show that n!nz Γ (z) = n z(z + 1)···(z + n) and we obtain Euler’s limit formula for the gamma function lim Γn(z) = Γ (z). n→∞ We write Γ (z) −Γn(z) = I1(z) + I2(z) + I3(z), where Z ∞ −t z−1 I1(z) = e t dt , n Z n n −t t z−1 I2(z) = e − 1 − t dt , n/2 n Z n/2 n −t t z−1 I3(z) = e − 1 − t dt . 0 n Obviously I1(z) tends to zero as n → ∞. For I2 and I3 we have that 0 ≤ t < n, and by the Taylor expansion of the logarithm we obtain t n t log 1 − = nlog 1 − = −t − L, n n where t2 1 1 t 1 t2 L = + + + ··· . n 2 3 n 4 n2 It follows that t n 0 < e−t − 1 − = e−t − e−Le−t ≤ e−t , n and thus I2(z) tends to zero as n → ∞.