More Properties of the Incomplete Gamma Functions

Home , Error function

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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 0 is mapped to (r, θ):0

√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 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). 2 2 a t√t a 2a Z − Also, b2t2 ∞ e π 2 2 (2.12) − dt = ea b erfc(ab). t2 + a2 2a Z0

6 References

[1] A. Apelblat (1983) Table of Definite and Infinite Integrals. Physical Sciences Data, Vol. 13, Elsevier Scientific Publishing Co., Amsterdam.

[2] Gradshteyn, I.S.; Ryzhik, I.M. (1980) Tables of Integrals, Series, and Products 4th edition. New York: Academic Press.

[3] Miller, Allen R.; Moskowitz, Ira S. (1998) On certain Generalized incomplete Gamma functions, J. Comp. Appl. Math. 91, No 2, 179-190.

[4] Temme, Nico. (1975) Uniform Asymptotic Expansions of the Incomplete Gamma Functions and the Incomplete Beta Function. Math. Comp. 29 No 132, 1109-1114.