The Beta Function

The area under \t^{p-1}(1-t)^{q-1}\ is the beta function — morph its shape with the \p\, \q\ sliders. Also covers \B(p,q) = \\Gamma(p)\\Gamma(q)/\\Gamma(p+q)\, the bridge to the gamma function.

B(p,q)=int01tp−1(1−t)q−1,dtquad(p,q>0)B(p,q)=\\int_0^1 t^{p-1}(1-t)^{q-1}\\,dt\\quad(p,q>0)
B(p,q)=fracGamma(p)Gamma(q)Gamma(p+q)B(p,q)=\\frac{\\Gamma(p)\\Gamma(q)}{\\Gamma(p+q)}

Part of these learning paths

Examples & Exercises

Prove that the improper integral B(p,q)=∫01tp−1(1−t)q−1 dtB(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}\,dt converges for p,q>0p, q > 0.

∫01/2tp−1(1−t)q−1 dt≍∫01/2tp−1 dt<∞  ⟺  p>0\int_0^{1/2} t^{p-1}(1-t)^{q-1}\,dt \asymp \int_0^{1/2} t^{p-1}\,dt < \infty \iff p > 0
p=0.5p = 0.5 — blows up at t=0t = 0, yet the integral converges.
Show proof
  1. Split the integral at t=1/2t = 1/2. On (0,1/2](0, 1/2] we have 12≤1−t<1\frac12 \leq 1-t < 1, so (1−t)q−1(1-t)^{q-1} is trapped between the positive constants c1=min⁡(1,21−q)c_1 = \min(1, 2^{1-q}) and c2=max⁡(1,21−q)c_2 = \max(1, 2^{1-q}). Hence c1tp−1≤tp−1(1−t)q−1≤c2tp−1c_1 t^{p-1} \leq t^{p-1}(1-t)^{q-1} \leq c_2 t^{p-1}, and the comparison test reduces the convergence of ∫01/2tp−1(1−t)q−1 dt\int_0^{1/2} t^{p-1}(1-t)^{q-1}\,dt to that of ∫01/2tp−1 dt\int_0^{1/2} t^{p-1}\,dt.
  2. ∫δ1/2tp−1 dt=(1/2)p−δpp\int_\delta^{1/2} t^{p-1}\,dt = \frac{(1/2)^p - \delta^p}{p} for p≠0p \neq 0, which converges to 2−pp\frac{2^{-p}}{p} as δ→0+\delta \to 0^+ when p>0p > 0. For p=0p = 0 it is −ln⁡(2δ)→∞-\ln(2\delta) \to \infty, and for p<0p < 0 it is δp∣p∣→∞\frac{\delta^p}{|p|} \to \infty. So ∫01/2tp−1 dt\int_0^{1/2} t^{p-1}\,dt converges exactly when p>0p > 0.
  3. The end t=1t = 1 is symmetric. With s=1−ts = 1-t, ∫1/21−εtp−1(1−t)q−1 dt=∫ε1/2(1−s)p−1sq−1 ds\int_{1/2}^{1-\varepsilon} t^{p-1}(1-t)^{q-1}\,dt = \int_{\varepsilon}^{1/2} (1-s)^{p-1} s^{q-1}\,ds (a legitimate C1C^1 substitution on a compact interval), so the same argument with pp and qq swapped shows ∫1/21tp−1(1−t)q−1 dt\int_{1/2}^1 t^{p-1}(1-t)^{q-1}\,dt converges exactly when q>0q > 0.
  4. On the middle piece [δ,1−δ][\delta, 1-\delta] the integrand is continuous, so the integral is always finite. Hence the improper integral B(p,q)B(p,q) converges exactly when p>0p > 0 and q>0q > 0 — the divergences at the two ends cannot cancel, since the integrand is positive.

Prove B(p,q)=B(q,p)B(p,q) = B(q,p) with the substitution u=1−tu = 1-t.

B(p,q)=∫01tp−1(1−t)q−1 dt=∫01uq−1(1−u)p−1 du=B(q,p)B(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}\,dt = \int_0^1 u^{q-1}(1-u)^{p-1}\,du = B(q,p)
p=2,q=3p = 2, q = 3 — B(2,3)=B(3,2)=1/12B(2,3) = B(3,2) = 1/12.
Show proof
  1. Problem 1 gives convergence for p,q>0p, q > 0, so we may substitute u=1−tu = 1-t (du=−dtdu = -dt) in the truncated integral ∫δ1−εtp−1(1−t)q−1 dt\int_\delta^{1-\varepsilon} t^{p-1}(1-t)^{q-1}\,dt (a C1C^1 substitution on a compact interval).
  2. The interval flips: t=δt = \delta gives u=1−δu = 1-\delta and t=1−εt = 1-\varepsilon gives u=εu = \varepsilon, so ∫δ1−εtp−1(1−t)q−1 dt=∫1−δε(1−u)p−1uq−1(−du)=∫ε1−δuq−1(1−u)p−1 du\int_\delta^{1-\varepsilon} t^{p-1}(1-t)^{q-1}\,dt = \int_{1-\delta}^{\varepsilon} (1-u)^{p-1}u^{q-1}(-du) = \int_{\varepsilon}^{1-\delta} u^{q-1}(1-u)^{p-1}\,du.
  3. Sending δ,ε→0+\delta, \varepsilon \to 0^+, convergence from problem 1 turns the two sides into B(p,q)B(p,q) and B(q,p)B(q,p), so B(p,q)=B(q,p)B(p,q) = B(q,p). Flipping the curve tp−1(1−t)q−1t^{p-1}(1-t)^{q-1} left-right leaves the area underneath unchanged.

Show B(p+1,q)=pqB(p,q+1)B(p+1,q) = \frac{p}{q}B(p,q+1) by parts, then deduce B(p+1,q)=pp+qB(p,q)B(p+1,q) = \frac{p}{p+q}B(p,q).

B(p+1,q)=pp+q B(p,q)B(p+1,q) = \frac{p}{p+q}\,B(p,q)
p=1,q=3p = 1, q = 3 — B(2,3)=14B(1,3)=112B(2,3) = \frac14 B(1,3) = \frac{1}{12}.
Show proof
  1. Apply parts to B(p+1,q)=∫01tp(1−t)q−1 dtB(p+1,q) = \int_0^1 t^p(1-t)^{q-1}\,dt with u=tpu = t^p, dv=(1−t)q−1 dtdv = (1-t)^{q-1}\,dt. Then du=ptp−1 dtdu = p t^{p-1}\,dt and v=−(1−t)qqv = -\frac{(1-t)^q}{q}.
  2. The boundary term [−tp(1−t)qq]01\left[-\frac{t^p(1-t)^q}{q}\right]_0^1 vanishes: at t=1t = 1, (1−t)q→0(1-t)^q \to 0 since q>0q > 0; at t=0t = 0, tp→0t^p \to 0 since p>0p > 0. Hence B(p+1,q)=pq∫01tp−1(1−t)q dt=pqB(p,q+1)B(p+1,q) = \frac{p}{q}\int_0^1 t^{p-1}(1-t)^q\,dt = \frac{p}{q}B(p,q+1).
  3. Meanwhile, multiplying the integrand of B(p,q)=∫01tp−1(1−t)q−1 dtB(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}\,dt by 1=t+(1−t)1 = t + (1-t) gives ∫01tp(1−t)q−1 dt+∫01tp−1(1−t)q dt\int_0^1 t^p(1-t)^{q-1}\,dt + \int_0^1 t^{p-1}(1-t)^q\,dt, that is, B(p+1,q)+B(p,q+1)B(p+1,q) + B(p,q+1).
  4. From the second step, B(p,q+1)=qpB(p+1,q)B(p,q+1) = \frac{q}{p}B(p+1,q); substituting gives B(p,q)=B(p+1,q)+qpB(p+1,q)=p+qpB(p+1,q)B(p,q) = B(p+1,q) + \frac{q}{p}B(p+1,q) = \frac{p+q}{p}B(p+1,q). Rearranging yields B(p+1,q)=pp+qB(p,q)B(p+1,q) = \frac{p}{p+q}B(p,q).
  5. A note (an idea-level sketch, not a proof): the bridge to the gamma function, B(p,q)=Γ(p)Γ(q)Γ(p+q)B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}, follows the standard sketch — write Γ(p)Γ(q)\Gamma(p)\Gamma(q) as a double integral over (0,∞)2(0,\infty)^2; the polar substitution t=r2sin⁡2θt = r^2\sin^2\theta, s=r2cos⁡2θs = r^2\cos^2\theta then separates the radial part into Γ(p+q)\Gamma(p+q) and the angular part into B(p,q)B(p,q). The full justification, including Fubini's theorem on the unbounded domain, goes beyond this scene, so we take the relation as the bridge, not as a proved theorem here.

10. Integration