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Integral of the fractional power kernel

Statement

For $β > 0$ and $t > 0$ , $\int_{0}^{t} (t - s)^{β - 1} d s = \frac{t^{β}}{β} .$ When $0 < β < 1$ the integral is improper at $s = t$ ; for $β \geq 1$ it is an ordinary integral. It converges in both cases because $B (β, 1)$ is defined for positive arguments.

Facts & Assumptions

Given: $β > 0$ and $t > 0$ .

[F1]

Euler's real Beta integral: $B (p, q) = \int_{0}^{1} x^{p - 1} (1 - x)^{q - 1} d x$ for $p, q > 0$ .

[F2]

The real Beta--Gamma identity: $B (p, q) = Γ (p) Γ (q) / Γ (p + q)$ for $p, q > 0$ .

[F3]

The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ : $Γ (x + 1) = x Γ (x)$ for $x > 0$ .

[F4]

The real Gamma function by Euler's integral: $Γ (1) = 1$ .

[F5]

Change of variable in an improper integral: a monotone differentiable change of variable preserves an improper integral when the transformed integral converges.

Proof

1.1givenF1F5construct

In the improper integral, set $x = (t - s) / t$ . On every compact truncation this is a differentiable decreasing substitution with $d s = - t d x$ , so [F5] and passage to the endpoint limits give $\int_{0}^{t} (t - s)^{β - 1} d s = t^{β} \int_{0}^{1} x^{β - 1} d x = t^{β} B (β, 1),$ where the last equality is [F1] because $(1 - x)^{0} = 1$ on $(0, 1)$ .

2.1step 1.1F2F3F4algebra∎

By [F2] and [F4], $B (β, 1) = Γ (β) / Γ (β + 1)$ . Applying [F3] gives $B (β, 1) = 1 / β$ . Substitution into step 1.1 proves the formula and convergence of the stated improper integral.

Depends on

Used by

Dependency tree · two levels

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Sources

William F. Trench, Introduction to Real Analysis, Theorem 3.4.13 (standard reference, not scraped)
University of Toronto MAT237Y1, The Gamma Function and the Beta Function, §2 (standard reference, not scraped)