MORE PROPERTIES OF THE INCOMPLETE GAMMA FUNCTIONS R. AlAhmad Mathematics Department, Yarmouk University, Irbid, Jordan 21163, email:rami [email protected] February 17, 2015 Abstract In this paper, additional properties of the lower gamma functions and the error functions are introduced and proven. In particular, we prove interesting relations between the error functions and Laplace transform. AMS Subject Classification: 33B20 Key Words and Phrases: Incomplete beta and gamma functions, error functions. 1 Introduction Definition 1.1. The lower incomplete gamma function is defined as: arXiv:1502.04606v1 [math.CA] 29 Nov 2014 x s 1 t γ(s, x)= t − e− dt. Z0 Clearly, γ(s, x) Γ(s) as x . The properties of these functions are listed in −→ −→ ∞ many references( for example see [2], [3] and [4]). In particular, the following properties are needed: Proposition 1.2. [1] s x 1. γ(s +1, x)= sγ(s, x) x e− , − x 2. γ(1, x)=1 e− , − 1 3. γ 2 , x = √π erf (√x) . 1 2 Main Results Proposition 2.1. [1]For a< 0 and a + b> 0 ∞ a 1 Γ(a + b) x − γ(b, x) dx = . − a Z0 Proposition 2.2. For a =0 6 √t s (ar)2 1 s +1 2 r e− dr = γ( , a t). 2as+1 2 Z0 Proof. The substitution u = r2 gives √ 2 t a t s−1 s (ar)2 1 u 1 s +1 2 r e− dr = e− u 2 du = γ( , a t). 2as+1 2as+1 2 Z0 Z0 Proposition 2.3. ξ x x 1 ηt η t − e− dt = γ(x,ηξ) Z0 Proof. ηξ t x 1 γ(x,ηξ)= e− t − 0 Z ξ ξ ηt x 1 x x 1 ηt = e− (ηt) − = η t − e− dt Z0 Z0 The incomplete gamma function γ(a, t) satisfies Lemma 2.4. For t 0, ≥ π/2 2 2 2a 1 2b 1 γ(a, t)= γ(a + b, t sec (θ)) cos − (θ) sin − (θ)dθ. Γ(b) Z0 Proof. The strip (x, y):0 <x< √t,y > 0 is mapped to (r, θ):0 <r< √t sec(θ), 0 < { } { θ<π/2 , where x = r cos(θ) and y = r sin(θ). } Now, Using the substitutions z = x2 and v = y2, we get √t t ∞ x2 2a 1 y2 2b 1 ∞ z a 1 v b 1 4 e− x − e− y − dydx = e− z − e− v − dvdz. Z0 Z0 Z0 Z0 2 This result gives t ∞ z a 1 v b 1 γ(a, t)Γ(b)= e− z − e− v − dvdz Z0 Z0 √t ∞ x2 2a 1 y2 2b 1 =4 e− x − e− y − dydx Z0 Z0 π/2 √t sec(θ) 2a 1 2b 1 r2 2a+2b 1 (2.1) =4 cos − (θ) sin − (θ)e− r − drdθ. Z0 Z0 Using Proposition 2.2 and Equation 2.1 we get π/2 2 2a 1 2b 1 γ(a, t)Γ(b)=2 γ(a + b, t sec (θ)) cos − (θ) sin − (θ)dθ. Z0 As a remark, in Lemma 2.4, if we let t then γ(a, t) Γ(a) and γ(a + −→ ∞ −→ b, t sec2(θ)) Γ(a + b). This proves the well-known result that is −→ π/2 Γ(a)Γ(b) 2a 1 2b 1 (2.2) β(a, b)= =2 cos − (θ) sin − (θ)dθ. Γ(a + b) Z0 As a manner of fact, one can easily show that It is Known that the gamma function satisfies the property Proposition 2.5. The gamma function Γ satisfies Γ(1 a)Γ(a)= π csc(πa). − The following is an extension of this result Lemma 2.6. For t 0, ≥ π/2 t sec2(θ) 2a 1 γ(a, t)Γ(1 a)=2 (1 e− )cot − (θ)dθ − 0 − Z π/2 t sec2(θ) 2a 1 =π csc(πa) 2 e− cot − (θ)dθ. − Z0 Proof. Proposition 1.2 implies 2 t sec2(θ) (2.3) γ(1, t sec (θ))=1 e− . − Then Lemma 2.4 with b =1 a gives the result. − This lemma gives the following 3 Proposition 2.7. For 1 <a< 0, − t ∞ 1 e− Γ(a + 1) − dt = . t1 a − a Z0 − Proof. Using Lemma 2.1 we get ∞ a 1 Γ(a + b) x − γ(b, x) dx = . − a Z0 Multiply both sides by Γ(1 b) and use Lemma 2.6 to get − π/2 Γ(1 b)Γ(a + b) ∞ t sec2(θ) a 1 2a 1 − =2 (1 e− )t − cot − (θ)dθdt. − a − Z0 Z0 Interchange the order of the integral and use the substitution u = t sec2(θ) to get: u π/2 Γ(1 b)Γ(a + b) ∞ 1 e− 2a+2b 1 1 2b − = − du 2 cos − (θ) sin − (θ)dθdθ . − a u1 a Z0 − Z0 ! Use Equation 2.2 to get the result. Corollary 2.8. The error function satisfies π/2 2 1 2 2b 1 erf(√at)= γ( + b, at sec (θ)) sin − (θ)dθ. √πΓ(b) 2 Z0 In particular, π/2 2 at sec2(θ) erf(√at)=1 e− dθ. − π Z0 Proof. By Lemma 2.4 with a =1/2 and Proposition 1.2, we get π/2 2 1 2 2b 1 √π erf(√at)= γ(1/2, at)= γ( + b, at sec (θ)) sin − (θ)dθ. Γ(b) 2 Z0 Therefore, π/2 2 1 2 2b 1 erf(√at)= γ( + b, at sec (θ)) sin − (θ)dθ. √πΓ(b) 2 Z0 The last equation with b = 1 gives 2 π/2 erf(√at)= γ(1, at sec2(θ))dθ. π Z0 Using Equation 2.3 implies π/2 π/2 2 2 2 at sec2(θ) erf(√at)= γ(1, at sec (θ))dθ =1 e− dθ. π − π Z0 Z0 i.e., π/2 2 at sec2(θ) (2.4) erfc(√at)=1 erf(√at)= e− dθ. − π Z0 4 Multiplying Equation 2.4 by eat implies π/2 at 2 at tan2(θ) (2.5) e erfc(√at)= e− dθ. π Z0 Equation 2.4 gives the following results Lemma 2.9. π/2 2 ∞ 2 2a ∞ F (s ) (2.6) f(t) erfc(a√t)dt = F (a2 sec2(θ))dθ = ds, 2 2 0 π 0 π a s√s a Z Z Z − where F (s)=( f)(s) is the Laplace transform of f(t). L Proof. Equation 2.4 implies π/2 ∞ 2 ∞ ta2 sec2(θ) f(t) erfc(a√t)dt = f(t)e− dθdt. π Z0 Z0 Z0 By interchanging the order of the integral, we get: π/2 ∞ 2 ∞ ta2 sec2(θ) (2.7) f(t) erfc(a√t)dt = f(t)e− dtdθ. π Z0 Z0 Z0 st Now, Since the Laplace transform is given as F (s)=( f)(s)= ∞ f(t)e− dt, then L 0 2 2 R ∞ ta sec (θ) 2 2 (2.8) f(t)e− dt = F (a sec (θ)). Z0 Then Equations 2.7 and 2.9 prove the result that π/2 ∞ 2 f(t) erfc(a√t)dt = F (a2 sec2(θ))dθ. π Z0 Z0 In addition, the substitution s = a sec2 θ implies that π/2 2 2 2a ∞ F (s ) F (a2 sec2(θ))dθ = ds. 2 2 π 0 π a s√s a Z Z − Corollary 2.10. For r > 1, − 3 ∞ Γ(r + ) 2aΓ(r + 1) ∞ 1 tr erfc(a√t)dt = 2 = ds. 2r+2 2r+3 2 2 0 a √π(1 + r) π a s √s a Z Z − Moreover, for µ> 1 − µ ∞ 1 Γ(1 + ) tµ erfc(at)dt = 2 . √π aµ+1(1 + µ) Z0 5 Proof. Lemma 2.9 implies that π/2 ∞ 2 Γ(r + 1) tr erfc(a√t)dt = dθ. π (a2 sec2 θ)r+1 Z0 Z0 Use Equation 2.2 to get that π/2 1 1 π/2 1 Γ(r + 3 )√π dθ = (cos2 θ)r+1dθ = 2 . (a2 sec2 θ)r+1 a2r+2 a2r+2 2Γ(r + 2) Z0 Z0 Therefore, 3 3 ∞ 2 Γ(r + )√π Γ(r + ) (2.9) tr erfc(a√t)dt = Γ(r + 1) 2 = 2 . π 2a2r+2Γ(r + 2) a2r+2√π(1 + r) Z0 Now, the substitution u = √t into equation 2.9 with a = 1 gives that 3 ∞ ∞ Γ(r + ) 2 u2r+1 erfc(u)du = tr erfc(√t)dt = 2 . √π(1 + r) Z0 Z0 Hence, µ 1 3 µ ∞ µ 1 Γ( −2 + 2 ) 1 Γ(1 + 2 ) u erfc(u)du = µ 1 = . 2 √π(1 + − ) √π 1+ µ Z0 2 Lastly, µ ∞ 1 ∞ 1 Γ(1 + ) tµ erfc(at)dt = tµ erfc(t)dt = 2 . aµ+1 √π aµ+1(1 + µ) Z0 Z0 Following similar proof of Lemma 2.9 and using Equation 2.5, we get the following result Lemma 2.11. π/2 2 ∞ 2 2 2a ∞ F (s ) (2.10) ea tf(t) erfc(a√t)dt = F (a2 tan2(θ))dθ = ds, π π s2 + a2 Z0 Z0 Za where F (s)=( f)(s) is the Laplace transform of f(t). L Corollary 2.12. For r > 1, − ∞ 2 2aΓ(r + 1) ∞ 1 trea t erfc(a√t)dt = ds. π s2r+2(s2 + a2) Z0 Za Substitute the Laplace transform of the dirac delta function f(t) = δ(t b) which is bs − F (s)= e− into Equations 2.6 and 2.10 to get that Proposition 2.13. b2t2 ∞ e π (2.11) − dt = erfc(ab).