The Gamma Function

The factorial stretched to real arguments — Γ(x+1)=xΓ(x)\Gamma(x+1) = x\Gamma(x), so Γ(n+1)=n!\Gamma(n+1) = n!. Watch the improper integral converge for x>0x > 0.

Γ(x)=∫0∞tx−1e−t dt(x>0)\Gamma(x) = \int_0^\infty t^{x-1} e^{-t}\,dt \quad (x > 0)
Γ(x+1)=x Γ(x)\Gamma(x+1) = x\,\Gamma(x)

Part of these learning paths

Examples & Exercises

Prove Γ(x+1)=xΓ(x)\Gamma(x+1) = x\Gamma(x) by integration by parts.

Γ(x+1)=∫0∞txe−t dt=x Γ(x)\Gamma(x+1) = \int_0^\infty t^x e^{-t}\,dt = x\,\Gamma(x)
x=2.5x = 2.5 — Γ(3.5)=2.5 Γ(2.5)\Gamma(3.5) = 2.5\,\Gamma(2.5).
Show proof
  1. Set u=txu = t^x, dv=e−t dtdv = e^{-t}\,dt, so du=xtx−1 dtdu = x t^{x-1}\,dt and v=−e−tv = -e^{-t}. Integration by parts gives Γ(x+1)=[−txe−t]0∞+x∫0∞tx−1e−t dt\Gamma(x+1) = \left[-t^x e^{-t}\right]_0^\infty + x \int_0^\infty t^{x-1} e^{-t}\,dt.
  2. The boundary term vanishes: since x>0x > 0, tx→0t^x \to 0 as t→0+t \to 0^+, and txe−t→0t^x e^{-t} \to 0 as t→∞t \to \infty (the exponential beats the polynomial). So [−txe−t]0∞=0\left[-t^x e^{-t}\right]_0^\infty = 0.
  3. Hence Γ(x+1)=x∫0∞tx−1e−t dt=x Γ(x)\Gamma(x+1) = x \int_0^\infty t^{x-1} e^{-t}\,dt = x\,\Gamma(x).

Show that Γ(1)=1\Gamma(1) = 1, and from it prove Γ(n+1)=n!\Gamma(n+1) = n!.

Γ(1)=∫0∞e−t dt=1,Γ(n+1)=n!\Gamma(1) = \int_0^\infty e^{-t}\,dt = 1, \quad \Gamma(n+1) = n!
x=1x = 1 — Γ(1)=1\Gamma(1) = 1.
Show proof
  1. Γ(1)=∫0∞e−t dt=[−e−t]0∞=0−(−1)=1\Gamma(1) = \int_0^\infty e^{-t}\,dt = \left[-e^{-t}\right]_0^\infty = 0 - (-1) = 1.
  2. Induction with the functional equation from the previous problem: Γ(1)=0!\Gamma(1) = 0! holds for n=0n = 0.
  3. Assuming Γ(n)=(n−1)!\Gamma(n) = (n-1)!, we get Γ(n+1)=n Γ(n)=n (n−1)!=n!\Gamma(n+1) = n\,\Gamma(n) = n\,(n-1)! = n!. Hence Γ(n+1)=n!\Gamma(n+1) = n! for every natural number nn.

Prove that the improper integral defining Γ(x)\Gamma(x) converges for every x>0x > 0.

∫01tx−1e−t dt≤∫01tx−1 dt=1x\int_0^1 t^{x-1}e^{-t}\,dt \le \int_0^1 t^{x-1}\,dt = \frac1x
x=0.5x = 0.5 — converges near 0 too.
Show proof
  1. Split the integral at t=1t = 1. On (0,1](0, 1], e−t≤1e^{-t} \leq 1 gives 0<tx−1e−t≤tx−10 < t^{x-1}e^{-t} \leq t^{x-1}. Since ∫01tx−1 dt=1x<∞\int_0^1 t^{x-1}\,dt = \frac1x < \infty exactly when x>0x > 0, the comparison test gives convergence of ∫01tx−1e−t dt\int_0^1 t^{x-1}e^{-t}\,dt.
  2. On [1,∞)[1, \infty), write tx−1e−t=(tx−1e−t/2) e−t/2t^{x-1}e^{-t} = \bigl(t^{x-1}e^{-t/2}\bigr)\,e^{-t/2}. With M=max⁡t≥1tx−1e−t/2<∞M = \max_{t \geq 1} t^{x-1}e^{-t/2} < \infty (continuous and tending to 00 as t→∞t \to \infty), we get tx−1e−t≤Me−t/2t^{x-1}e^{-t} \leq M e^{-t/2}.
  3. Since ∫1∞e−t/2 dt=2<∞\int_1^\infty e^{-t/2}\,dt = 2 < \infty, the comparison test gives convergence of ∫1∞tx−1e−t dt\int_1^\infty t^{x-1}e^{-t}\,dt. Both pieces converge, so Γ(x)\Gamma(x) is well-defined for every x>0x > 0.

10. Integration