1. Gamma Function We will prove that the improper integral Z ∞ Γ(x) = e−ttx−1dt 0 exists for every x > 0. The function Γ(x) is called the Gamma function. Let us recall the comparison test for improper integrals. Theorem 1.1. (Comparison Test for Improper Integral of Type I) Let f(x), g(x) be two continuous functions on [a, ∞) such that 0 ≤ f(x) ≤ g(x) for all x ≥ a. Then Z ∞ Z ∞ (1) If g(x)dx is convergent, so is f(x)dx. a a Z ∞ Z ∞ (2) If f(x)dx is divergent to infinity, so is g(x)dx. a a Theorem 1.2. (Limit Comparison Test) Let f(x), g(x) be two nonnegative continuous functions on [a, ∞). Suppose that f(x) lim = L with L > 0. x→∞ g(x) Z ∞ Z ∞ Then both g(x)dx and f(x)dx are convergent or divergent. a a Z ∞ Lemma 1.1. For every s > 0, the improper integral e−stdt converges. 0 Proof. Let us compute Z N e−sM − 1 e−stdt = . 0 −s By definition, Z ∞ Z N e−sM − 1 1 e−stdt = lim e−stdt = lim = . 0 N→∞ 0 N→∞ −s s The limit exists and hence the improper integral converges. Now let us study the case when x ≥ 1.
Lemma 1.2. Let n be a natural number. Then tn−1 lim = 0. 1 t t→∞ e 2 Proof. By L’ Hospital rule, tn−1 (n − 1)tn−2 lim 1 = lim 1 . t→∞ 2 t t→∞ 1 2 t e 2 e Since tn−1 is a polynomial of degree n − 1, we know dn tn−1 = 0. dtn Inducitvely, we find tn−1 0 lim 1 = lim n 1 = 0. t→∞ 2 t t→∞ 1 2 t e 2 e 1 2
By the definition of limit, we choose = 1, there exists M > 0 such that for all t ≥ M tn−1 < = 1. 1 t e 2
n−1 1 t Hence for t ≥ M, 0 ≤ t < e 2 . This implies that for t ≥ M,
−t n−1 −t 1 t − 1 t (1.1) 0 ≤ e t ≤ e · e 2 = e 2 . Z ∞ − 1 t Lemma 1.1 implies that e 2 dt is convergent. (1.1) and Comparison test implies that 0 Z ∞ −t n−1 e t dt is convergent for every n ∈ N. 0 Let x ≥ 1 be any real number. Let [x] be the largest integer so that [x] ≤ x < [x] + 1. Then for t ≥ 0, (1.2) 0 ≤ e−ttx−1 ≤ e−tt[x]. Z ∞ Z ∞ Since e−tt[x]dx is convergent, by comparison test and (1.2), we find e−ttx−1dt is 0 0 convergent. Now let us study the case when 0 < x < 1. Then we know 1 tx−1 t ≤ ≤ . 1 t 1 t 1 t e 2 e 2 e 2 We know that t 1 lim = lim = 0. 1 t 1 t t→∞ e 2 t→∞ e 2 By the Sandwich principle, tx−1 lim = 0 1 t t→∞ e 2 for 0 < x < 1. (Of course, this statement is true for all x > 0.) This implies that
−t x−1 − 1 t 0 ≤ e t ≤ e 2 , t ≥ 1. Z ∞ By comparison test, e−ttx−1dt is convergent for 0 < x < 1. Now, we need to verify that 1 Z 1 Z 1 t −t x−1 e e t dt = 1−x dt 0 0 t −t Z 1 e −t x−1 is convergent. Notice that lim 1−x = ∞. Hence the integral e t dt is a type II t→0 t 0 improper integral. Theorem 1.3. Let f, g ∈ C(a, b] and 0 ≤ f(x) ≤ g(x) for all a < x ≤ b. Z b Z b (1) If g(x)dx is convergent, so is f(x)dx. a a Z b Z b (2) If f(x)dx is divergent to infinity, so is g(x)dx. a a 3
Note that 0 < e−ttx−1 ≤ etx−1. We know that Z 1 Z 1 tx−1dt = lim tx−1dt 0 b→0 b x 1 t = lim b→0 x b 1 bx = lim − b→0 x x 1 = . x Z 1 By the comparison test, e−ttx−1dt is convergent. 0 Theorem 1.4. For every x > 0, the improper integral Z ∞ Γ(x) = e−ttx−1dt 0 is convergent. Proposition 1.1. For x > 0, Γ(x + 1) = xΓ(x). Proof. For every N > 0, using integration by parts, Z N Z N −t x x −t N −t x−1 e t dt = −t e 0 + x e t dt 0 0 Z N = −N xe−N + x e−ttx−1dt. 0 N x We have seen that lim N xe−N = lim = 0 for all x > 0. Then N→∞ N→∞ eN Z N Γ(x + 1) = lim e−ttxdt N→∞ 0 Z N = lim (−N xe−N + x e−ttx−1dt) N→∞ 0 Z N = x lim e−ttx−1dt N→∞ 0 = xΓ(x). Corollary 1.1. For every n ≥ 0, Γ(n + 1) = n!. Z ∞ Proof. We know e−tdt = 1. Hence Γ(1) = 1. We can prove the formula by induction. 0 Assume that the statement is true for some nonnegative integer k, i.e. Γ(k + 1) = k!. By the previous proposition, Γ(k + 2) = Γ((k + 1) + 1) = (k + 1)Γ(k + 1) = (k + 1) · k! = (k + 1)!. 4
Proposition 1.2. For any s > 0, x > 0, we have Z ∞ −st x−1 Γ(x) e t dt = x . 0 s Proof. This formula can be proved by using substitution u = st. Example 1.1. Z ∞ −st 2 2 3 10 e (2 − 3t + 5t )dt = − 2 + 3 . 0 s s s This can be proved by the previous proposition.
Here comes a question, if f(t) is a continuous function on [0, ∞) such that Z ∞ −st 2 3 10 e f(t)dt = − 2 + 3 , 0 s s s what can you say about f(t)? Let us introduction the notion of Laplace transformation. 5
2. Laplace Transformation, its inverse transformation, and linear homogeneous o.d.e of constant coefficients Z ∞ Definition 2.1. Let f(t) be a continuous function on [0, ∞). Suppose e−stf(t)dt con- 0 verges for s ≥ a for some a ∈ R. Denote Z ∞ L(f)(s) = e−stf(t)dt 0 called the Laplace transform of f.
Now, let us assume that the Laplace transform of f1, f2 exist on some domain D ⊂ R where f1, f2 are continuous functions on [0, ∞). For any real numbers a1, a2, we know that for any s ∈ D, Z ∞ −st L(a1f1 + a2f2)(s) = (a1f1(t) + a2f2(t))e dt 0 Z M −st = lim (a1f1(t) + a2f2(t))e dt M→∞ 0 Z M Z M −st −st = lim a1 f1(t)e dt + a2 f2(t)e dt M→∞ 0 0 Z M Z M −st −st = a1 lim f1(t)e dt + a2 lim f2(t)e dt M→∞ 0 M→∞ 0 = a1L(f1)(s) + a2L(f2)(s).
Proposition 2.1. Suppose that f1, f2 are two continuous functions on [0, ∞) such that their Laplace transform exist on some domain D ⊂ R for s. Then
L(a1f1 + a2f2)(s) = a1L(f1)(s) + a2L(f2)(s), s ∈ D, for any a1, a2 ∈ R. (We say that the Laplace transform is a linear transformation.)
Theorem 2.1. Suppose f1, f2 are two continuous functions on [0, ∞) such that their Laplace transform exist. If L(f1) = L(f2), then f1 = f2. Hence if g(s) = L(f)(s), we denote f(t) = L−1(g)(t) called then Laplace inverse transform. Proposition 2.2. Suppose f(t) is a smooth function on [0, ∞)(f (k)(t) exists for all k ≥ 1. Set f (0)(t) = f(t).) Assume that lim f (k)(t)e−st = 0 for all k ≥ 0. Then t→∞ L(f 0)(s) = −f(0) + sL(f)(s). Proof. Using integration by parts, Z N Z N −st 0 −st N −st e f (t)dt = f(t)e 0 + s e f(t)dt 0 0 Z N = f(N)e−sN − f(0) + s e−stf(t)dt. 0 6
By assumption, lim f(N)e−sN = 0 (consider k = 0.) Hence N→∞ Z ∞ Z N e−stf 0(t)dt = lim e−stf 0(t)dt 0 N→∞ 0 Z N = lim f(N)e−sN − f(0) + s e−stf(t)dt N→∞ 0 Z N = −f(0) + s lim e−stf(t)dt N→∞ 0 Z ∞ = −f(0) + s e−stf(t)dt. 0 Corollary 2.1. Under the same assumption as above, we have X L(f (n))(s) = snF (s) − sn−k−1f (k). k=0 When n = 2, we have L(f 00)(s) = s2F (s) − sf 0(0) − f(0). Now let us use Laplace transform to solve ordinary differential equation with constant coefficients. Example 2.1. Solve for y0 + 2y = 0 with initial condition y(0) = 1. Take Laplace transform, we obtain L(y0 + 2y)(s) = L(y0)(s) + 2L(y)(s) = 0. Let g(s) = L(y)(s). Using the previous proposition, L(y0)(s) = −y(0) + sL(y)(s) = −1 + sg(s). 1 Hence (−1 + sg(s)) + 2g(s) = 0. This implies that g(s) = . Then s + 2 y = L−1(g)(t) = e−2t. Example 2.2. Solve for y00 + y = 0 with initial condition y(0) = a and y0(0) = b. Let g(s) be the Laplace transform of y. Then L(y00)(s) + g(s) = 0. On the other hand, L(y00)(s) = s2g(s) − sy(0) − y0(0) = s2g(s) − as − b. This implies that s 1 (s2 + 1)g(s) = as + b =⇒ g(s) = a + b . s2 + 1 s2 s 1 We know = L(cos t)(s), and = L(sin t)(s). Hence y(t) = a cos t + b sin t. s2 + 1 s2 + 1 In general, we can solve for the linear (homogeneous) differential equation of constant coefficients through Laplace transformation. A linear homogeneous differential equation of constant coefficients is a differential equation (n) (n−1) 0 (2.1) any + an−1y + ··· + a1y + a0y = 0, where a0, ··· , an ∈ R. If an 6= 0, we say that the equation is of order n. Let us study the case when n = 2 : (2.2) ay00 + by0 + cy = 0. 7
Here a 6= 0. Taking the Laplace transformation of the equation, we have a(s2F (s) − sy0(0) − y(0)) + b(sF (s) − y(0)) + cF (s), where F (s) = L(y)(s). Therefore F (s) is a rational function in s and given by As + B F (s) = . as2 + bs + c where A = ay0(0) and B = ay0(0) + by(0). We can use partial fraction expansion to decom- pose F (s). Definition 2.2. The polynomial χ(s) = as2 + bs + c is called the characteristic polynomial of (2.2). We assume that a = 1 and let D = b2 − 4c be the discriminant of the characteristic polynomial. case 1: If D > 0, χ(s) has two distinct real roots. We denote the root of χ(s) by λ1 and λ2. Thus we may write 1 1 F (s) = C1 + C2 s − λ1 s − λ2 for some C1,C2 ∈ R. We know that 1 L(eat)(s) = . s − a Using the inverse transform and the linearity of L−1, we see that y(t) = L−1(F )(t) −1 1 −1 1 = C1L + C2L s − λ1 s − λ2 λ1t λ2t = C1e + C2e . Hence y is a linear combination of eλ1t and eλ2t. case 2: If D = 0, χ(s) has one repeated root, called λ. Then we write 1 1 F (s) = C + C . 1 s − λ 2 (s − λ)2 Recall that Z ∞ n λt −st n tλ Γ(n + 1) n! L(t e )(s) = e t e dt = n+1 = n+1 . 0 (s − λ) (s − λ) Thus we obtain y(t) = L−1(F )(t) 1 1 = C L−1 + C L−1 1 s − λ 2 (s − λ)2 λt λt = C1e + C2te λt = (C1 + C2t)e . case 3: If D < 0, by completing the square, we write χ(s) = (s − α)2 + β2. We write As B s − α β F (s) = + = C + C . (s − α)2 + β2 (s − α)2 + β2 1 (s − α)2 + β2 2 (s − α)2 + β2 8
Notice that s − α β L(eαt cos βt)(s) = , L(eαt sin βt) = . (s − α)2 + β2 (s − α)2 + β2 Thus y(t) = L−1(F )(t) s − α β = C L−1 + C L−1 1 (s − α)2 + β2 2 (s − α)2 + β2 αt αt = C1e cos βt + C2e sin βt αt = e (C1 cos βt + C2 sin βt). Remark. Let V be the subset of C2[0, ∞) consisting of y such that (2.2) holds. The above observation implies that V is in fact a two dimensional real vector (sub)space (of C2[0, ∞)).
This method can be applied to (2.1) for any n ∈ N. Definition 2.3. The characteristic polynomial of (2.1) is defined to be n χ(s) = ans + ··· + a1s + a0. Taking the Laplace transformation of (2.1) and using Corollary 2.1, we find that P (s) F (s) = χ(s) for some polynomial P (s) ∈ R[s]. Thus F (s) is a rational function and we can solve for y using the partial fraction for F (s). Thus we reduce our problems to the cases when χ(s) = (s − a)n or χ(s) = ((s − α)2 + β2)m. When χ(s) = (s − a)n, we write A A A F (s) = 1 + 2 + ··· + n s − a (s − a)2 (s − a)n 1 1! (n − 1)! = C + C + ··· + C , 0 s − a 1 (s − a)2 n−1 (s − a)n where Ci−1 = Ai/(i − 1)! for i = 1, ··· , n. By taking the inverse transform, we obtain n−1 at y(t) = (C0 + C1t + ··· + Cn−1t )e . When χ(s) = ((x − α)2 + β2)m, we write A s + B A s + B F (s) = 1 1 + ··· + m m (x − α)2 + β2 ((s − α)2 + β2)m C (s − a) + D β C (s − a) + D β = 1 1 + ··· + m m . (x − α)2 + β2 ((s − α)2 + β2)m
Here Ci,Dj are constants. By finding the inverse transform of (Ci(s − a) + Diβ)/(((s − α)2 + β2)i), we obtain y. 9
3. Beta Function In this section, x, y are positive real numbers. Let us define Γ(x)Γ(y) B(x, y) = . Γ(x + y) The function B(x, y) is called the Gamma function. It follows immediately from the defi- nition that B(x, y) = B(y, x), ∀x, y > 0. Using the property of Gamma function: Γ(x + 1) = xΓ(x) for x > 0, we obtain: Proposition 3.1. Suppose p, q > 0. Then (1) B(p, q) = B(p + 1, q) + B(p, q + 1). q q (2) B(p, q + 1) = B(p + 1, q) = B(p, q). p p + q Theorem 3.1. For x, y > 0, Z ∞ tx−1 B(x, y) = x+y dt. 0 (1 + t) Let us postpone the proof of this equation.
Let us consider the substitution t = tan2 θ with 0 ≤ θ ≤ π/2. Then dt = 2 tan θ sec2 θdθ. Using sec2 θ = tan2 θ + 1, Beta function B(x, y) can be rewritten as Z π 2 x−1 2 (tan θ) 2 B(x, y) = 2 x+y · 2 tan θ sec θdθ 0 (1 + tan θ) Z π 2x−2 2 tan θ 2 = 2 2x+2y · tan θ sec θdθ 0 sec θ Z π 2 2x−1 1 = 2 tan θ · 2x+2y−2 dθ 0 sec θ π 2x−1 Z 2 sin θ = 2 · cos2x+2y−2 θdθ 0 cos θ π Z 2 = 2 sin2x−1 θ cos2y−1 θdθ. 0 Using B(y, x) = B(x, y), we also obtain π Z 2 B(x, y) = 2 cos2x−1 θ sin2y−1 θdθ. 0 This is the second form of Beta function. Consider substitution t = sin2 θ for 0 ≤ θ ≤ π/2. Then dt = 2 sin θ cos θdθ, we obtain the third form of the Beta function: Z 1 B(x, y) = tx−1(1 − t)y−1dt. 0 We conclude that Theorem 3.2. The Beta function has the following forms: Z ∞ tx−1 (1) B(x, y) = x+y dt, 0 (1 + t) 10
π Z 2 (2) B(x, y) = 2 cos2x−1 θ sin2y−1 θdθ, 0 Z 1 (3) B(x, y) = tx−1(1 − t)y−1dt. 0 ∞ Z 2 Example 3.1. Compute e−x dx. 0 − 1 t 2 dt Let us make a change of variable t = x2. Then dt = 2xdx and thus dx = .The 2 integral can be rewritten as Z ∞ Z ∞ 1 −x2 1 −t − 1 Γ 2 e dx = e t 2 dt = . 0 2 0 2 Now, we only need to compute Γ(1/2). Using Beta function, 1 2 2 1 1 Γ 2 1 B , = 1 1 = Γ 2 2 Γ 2 + 2 2 On the other hand, Z π Z π 1 1 2 2· 1 −1 2· 1 −1 2 B , = 2 cos 2 θ sin 2 θdθ = 2 dθ = π. 2 2 0 0 √ Hence Γ(1/2) = π. We obtain that ∞ √ Z 2 π e−x dx = . 0 2 Z 2π Example 3.2. Compute sin4 θdθ. 0 We know Z 2π Z π Z π 4 2 4 2 2· 5 −1 2· 1 5 1 sin θdθ = 4 sin θdθ = 2 · 2 sin 2 θ cos 2 θdθ = 2B , . 0 0 0 2 2 We compute 5 1 Γ 5 Γ 1 B , = 2 2 . 2 2 Γ(3) Using Γ(x + 1) = xΓ(x), we obtain 5 3 1 1 3√ Γ = · · Γ = π. 2 2 2 2 4 √ √ 5 1 3 π · π 3 Z 2π 3 Hence B , = 4 = π. Thus sin4 θdθ = π. 2 2 2! 8 0 4 Now, let us go back to the proof of Theorem 3.1.
Z ∞ Z ∞ Γ(x)Γ(y) = e−ttx−1dt e−ssx−1ds 0 0 Z ∞ Z ∞ = e−(t+s)tx−1sy−1dtds. 0 0 11 Z ∞ Let us compute e−(t+s)tx−1dt. Consider t = su, we rewrite 0 Z ∞ Z ∞ e−(t+s)tx−1dt = sx e−(u+1)sux−1du. 0 0 Hence the integral becomes Z ∞ Z ∞ Γ(x)Γ(y) = e−(u+1)ssx+y−1ds ux−1du 0 0 Z ∞ Γ(x + y) x−1 = x+y · u du 0 (u + 1) Z ∞ ux−1 = Γ(x + y) x+y du. 0 (1 + u) Here we use Proposition 1.2. This shows that Γ(x)Γ(y) Z ∞ ux−1 = x+y du. Γ(x + y) 0 (1 + u) Z 3 Example 3.3. Compute (x − 1)10(x − 3)3dx. 1 Let t = (x − 1)/2. Then the integral becomes Z 3 Z 1 Z 1 (x − 1)10(x − 3)3dx = (2t)10(2t − 2)32dt = −214 t10(1 − t)3dt. 1 0 0 We obtain Z 3 10!3! (x − 1)10(x − 3)3dx = −214B(11, 4) = −214 · . 1 14!