A) the Gamma Function Is Magic B) the Method of Analytic Continuation Is

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The Gamma function My reasons for writing these notes: a) The Gamma function is magic b) The method of analytic continuation is very important in complex function theory c) The idea of subtracting divergent terms is very important in the theory of distributions (Chapter 5) d) The asymptotic evaluation of integrals is a very important ap- plied math (and other math) technique (Chapter 11)

1. Definition Deﬁnition For Rez > 0, set Z ∞ (1) Γ(z) = tz−1e−t dt. 0 tz−1 = e(z−1) log t, and the branch of the logarithm is taken to be − π < arg t ≤ π. The branch cut lies along the negative axis. Writing t = reiθ and z = x + iy, we have |tz−1| ≤ rx−1e−yθ, |e−t| = e−x, and for every δ > 0, the integral converges absolutely for x = Rez ≥ δ. By estimating the diﬀerence quotient Γ(z) − Γ(z ) Z ∞ tz−1 − tz0−1 0 = e−t dt, z − z0 0 z − z0 or by appealing to a general theorem, one can show that Γ(z) is analytic for z in the open right half plane. It diverges when z = 0, since the integrand is ∼ 1/t.

2. Factorial identity Integrate Z ∞ Γ(z + 1) = tze−t dt 0 by parts (u = tz, v = e−t). The result is (2) Γ(z + 1) = zΓ(z). Since Γ(1) = 1, one gets Γ(1 + 1) = 1Γ(1), Γ(2 + 1) = 2Γ(2), or (2) Γ(n + 1) = n!.

3. Value of Γ(1/2) and related integrals

Take z = 1/2 in (1) and make the change of variables t = u2. Then Z ∞ √ Γ(1/2) = 2 e−u2 du = π. 0 √ Then Γ(3/2) = Γ(1/2 + 1) = (1/2) π, etc. Example. Set t = au2 to get the formula Z ∞ p −au2 1 u e du = (p+1)/2 Γ((p + 1)/2). 0 2a One then ﬁnds, for example, ∞ ∞ Z 2 n! Z 2 1 u2ne−u du = , u2n+1e−u du = Γ((2n + 1)/2). 0 2 0 2

Example. Evaluate Z ∞ Z R (3) eix3 dx = lim eix3 dx. 0 R→∞ 0 Solution Try to integrate eiz3 around a suitable contour. The interval [0,R] on the real axis should be part of the contour. The function has no poles, hence no residues, and the answer will have to come from the contour integral alone. First, for z = R exp(iθ), we have z3 = R3 exp(3iθ). The real part is R3 exp(−3 sin θ). The exponent must be negative, so 0 ≤ 3 sin θ ≤ π. That means 0 ≤ θ ≤ π/3. Try the wedge of angle π/3 of the circle of radius R. The integral from x = 0 to x = R is the one we want. The integral along the arc z = R exp iθ, θ ∈ [0, π/3] will tend to zero as R → ∞, by Jordan’s Lemma. Finally, return to z = 0 along the ray z = r exp(iπ/3). That integral is Z 0 Z R (4) expi(reiπ/3)3eiπ/3 dr = −eiπ/3 e−ir3 dr. R 0 By Cauchy, Z R Z R eix3 dx − eiπ/3 e−ir3 dr = 0. 0 0 If you compare real and imaginary parts, you ﬁnd that √ Z Z 3 sin x3 = cos x3, which gets you nowhere. One has to take the wedge with angle π/6 instead. Then (4) becomes

Z 0 Z R expi(reiπ/6)3eiπ/6 dr = −eiπ/6 e−r3 dr. R 0 Set u = r3 in the integral, to get

Z R3 iπ/6 1 −2/3 −u −e 3 u e du. 0 The limit as R → ∞ is iπ/6 Z ∞ iπ/6 e 1 −1 −u e 1 − u 3 e du = − Γ( ). 3 0 3 3 Hence ∞ iπ/6 Z 3 e 1 eix dx = Γ( ). 0 3 3 (This integral is a special case of the one in Exercise 23 of the second complex variable exercise set.)

Example

Z 1 Z 1 2γ n Y Y α−1 β−1 1 n ··· (xj − x`) x (1 − xj) dx . . . dx j 0 0 1≤j<`≤n j=1 n−1 Y Γ(1 + γ + jγ)Γ(α + jγ)Γ(β + jγ) = Γ(1 + γ)Γ(α + β + (n + j − 1)γ j=0

This formula appears in the theory of random matrices. Its proof takes at least ﬁve pages. 4. Analytic continuation

Often, one has an analytic function deﬁned, in some domain D of C, by a Taylor series, Laurent series, or an integral (like the Gamma function). The function exists outside of D as well, but the representation is not valid. For example,

∞ 1 2 X 2 n (5) = − zn. z + 3 3 3 2 0 The series converges for |z| < 3/2, but the function is analytic for all z 6= −3/2. Given only the Taylor series and not the formula, one can enlarge the domain of definition. Replace z in the series by (z − 1) + 1, expand all the powers [(z − 1) + 1]n, and collect powers of (z − 1). One can P n prove that this new series, which has the form an(z − 1) , converges for |z − 1| < 5/2. (The distance from z = 1 to the pole z = −3/2 is 5/2). The new series, by construction, coincides with the old one in the overlap of |z| < 3/2 and |z − 1| < 5/2. The Gamma function, originally defined for Rez > 0 by the integral (1), is actually analytic in C − {0, −1, −2, −3,...}. One can see this in several ways. a) In (2), Γ(z+1) = zΓ(z), take z with −1 < Rez < 0, e.g. z = −1/2. Then Γ(1/2) √ Γ(−1/2) = = −2 π −1/2 is the only sensible value of Γ(−1/2). Similarly, one defines Γ(z) for −1 < Rez < 0, then for −2 < Rez < −1, and so on. b) There is a remarkable formula, with a simple but very clever proof: π Γ(z)Γ(1 − z) = , Rez > 0. sin πz √ One can find Γ(−1/2) from this (using Γ(3/2) = π/2). By letting z → 0, one sees that Γ(z) ∼ 1/z. So the residue at z = 0 is 1. This formula, like the one in a), can be used over and over again to extend Γ(z) to the left half plane. c) This method will be important, in other settings, in Chapter 5: Subtract terms that cause divergence. Write (1) as Z 1 Z ∞ + . 0 1 The second integral converges for all z, since e−t decreases faster at ∞ R 1 than any power of t. Now work on 0 . Write it as Z ∞ −t 2 3 n n z+n e − (1 − t + t /2! − t /3! + ··· (−1) t /n!) t n+1 dt 0 t n X Z 1 tz−1+k + (−1)k dt. k! k=0 0 I have subtracted the first n + 1 terms of the Taylor series for e−t. The numerator on the left starts with tn+1, so the fraction is finite, hence integrable, as t → 0. The integral therefore converges for Re(z + n) > −1, or z > −n−1. Of course, the subtracted term must be added back in. That is the second sum, which can be integrated. Now let n → ∞, R ∞ and combine with the 1 integral, to get the representation ∞ X (−1)k 1 Z ∞ Γ(z) = + tz−1e−1 dt. k! z + k n=0 1 This shows that Γ has a continuation that is analytic in C, except for poles at 0, −1, −2, −3,..., with residues (−1)k/k!. R ∞ k This method will be used to give meaning to integrals like −∞ φ(x)/x dx and other singular expressions, in Chapter 5.

5. Asymptotics The integral for Gamma can be used to get an asymptotic approx- imation for n!. I hope to treat the asymptotic evaluation of integrals more systematically in the chapter on Fourier integrals and in Chap- ter 11. The object is to approximate Γ(x + 1) as x → +∞. Write Z ∞ Z ∞ Z ∞ Γ(x+1) = txe−t dt t==xu xx+1 uxe−xu du = xx+1 ex(ln u−u) du. 0 0 0 The exponent is ≤ −1, and it is = −1 when u = 1. The main contri- bution to the integral when x is large will come from a neighborhood of u = 1, where the exponent is least negative. This statement can be made precise, but I won’t attempt to do so. The technique is known as Laplace’s method. Put xx+1 aside for a moment. Approximate ln u−u ∼ −1−(u−1)2/2, pick a small , and break the integral into a part 1 − ≤ u ≤ 1 + and the rest. One can show that “the rest” decays faster than the [1 − , 1 + ] integral and can be neglected by comparison. This leaves Z 1+ Z e−x e−(u−1)2/2 du u=1+=v e−x e−xv2/2 dv 1− √ − √ r r v= 2/xw Z x/2 −x 2 −w2 x→∞ −x 2π = e √ e dw = e . x − x/2 x Putting the xx+1 back in, we get √ Γ(x + 1) ∼ 2πx xxe−x For x = n, using Γ(n + 1) = n!, one gets Stirling’s formula √ n! ∼ 2πn nne−n. This is to be interpreted as n! lim √ = 1. n→∞ 2πn nne−n The diﬀerence √ n −n n! − 2πn n e tends to ∞ very rapidly. E.g., when n = 25: 25! = 1.551 × 1025, the ratio is 1.003, and the diﬀerence is 5.162 × 1022. There is also an elementary proof that uses only calculus and extreme cleverness.