The Gamma Function

The factorial stretched to real arguments. Values that jumped like stairs become one smooth curve.

The key rule: shifting by one multiplies by the shift amount. That rule fills in the values between integers. An integral, an area, draws the whole curve.

For example, picture a 🪜 staircase jumping 1, 2, 6, 24, with a smooth slide fitted between its steps. Now the heights between steps can be measured too.

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) for x>0x > 0 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. The integral is improper, so integrate by parts first on [ε,R][\varepsilon, R] with 0<ε<R<∞0 < \varepsilon < R < \infty.

    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}.

    Then ∫εRtxe−t dt=[−txe−t]εR+x∫εRtx−1e−t dt\int_{\varepsilon}^{R} t^x e^{-t}\,dt = \left[-t^x e^{-t}\right]_{\varepsilon}^{R} + x \int_{\varepsilon}^{R} t^{x-1} e^{-t}\,dt.

  2. In the limits ε→0+\varepsilon \to 0^+ and R→∞R \to \infty the boundary term vanishes: since x>0x > 0, εxe−ε→0\varepsilon^x e^{-\varepsilon} \to 0, and Rxe−R→0R^x e^{-R} \to 0 (the exponential beats the polynomial). So [−txe−t]εR→0\left[-t^x e^{-t}\right]_{\varepsilon}^{R} \to 0, and both integrals converge to the improper integrals.

  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.

    The base case is Γ(1)=1=0!\Gamma(1) = 1 = 0! at n=0n = 0; equivalently the induction for Γ(n)=(n−1)!\Gamma(n) = (n-1)! starts at n=1n = 1 with Γ(1)=0!=1\Gamma(1) = 0! = 1.

    The induction step below is only for n≥1n \ge 1.

  3. For n≥1n \ge 1, assuming Γ(n)=(n−1)!\Gamma(n) = (n-1)!, the functional equation applies at x=n≥1>0x = n \ge 1 > 0, so Γ(n+1)=n Γ(n)=n (n−1)!=n!\Gamma(n+1) = n\,\Gamma(n) = n\,(n-1)! = n! (the step is not applied at n=0n = 0, where neither the functional equation at x=0x = 0 nor (−1)!(-1)! is defined). Hence Γ(n+1)=n!\Gamma(n+1) = n! for every n≥0n \ge 0.

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}.

    The function g(t)=tx−1e−t/2g(t) = t^{x-1}e^{-t/2} is continuous on [1,∞)[1, \infty) and tends to 00 as t→∞t \to \infty, hence is bounded: choose RR with g(t)<1g(t) < 1 for t>Rt > R; on [1,R][1, R] it attains a maximum by the extreme value theorem, and beyond RR it is below 11.

    Since gg takes positive values, this maximum is attained, so with M=max⁡t≥1tx−1e−t/2<∞M = \max_{t \geq 1} t^{x-1}e^{-t/2} < \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=2e−1/2≈1.213<∞\int_1^\infty e^{-t/2}\,dt = 2e^{-1/2} \approx 1.213 < \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.

11. Integration