Dirichlet integral over positive octant region

This is Dirichlet's integral for the generalised region (x/a)^p + (y/b)^q + (z/c)^r ≤ 1 with x, y, z ≥ 0.

Dirichlet's standard result

For u, v, w ≥ 0 with u + v + w ≤ 1:

∭ u^(α−1) v^(β−1) w^(γ−1) du dv dw = Γ(α) Γ(β) Γ(γ) / Γ(α + β + γ + 1).

Substitution

Put (x/a)^p = u, (y/b)^q = v, (z/c)^r = w. Then the region becomes u + v + w ≤ 1, and

x = a u^(1/p)   →  dx = (a/p) u^(1/p − 1) du
y = b v^(1/q)   →  dy = (b/q) v^(1/q − 1) dv
z = c w^(1/r)   →  dz = (c/r) w^(1/r − 1) dw

So x^(l−1) dx = a^(l−1) u^((l−1)/p) · (a/p) u^(1/p − 1) du = (a^l/p) u^(l/p − 1) du. Similarly for y and z.

The integral

I = (a^l b^m c^n)/(p q r) ∭ u^(l/p − 1) v^(m/q − 1) w^(n/r − 1) du dv dw

Applying Dirichlet's result with α = l/p, β = m/q, γ = n/r:

I = (a^l b^m c^n)/(p q r) · Γ(l/p) Γ(m/q) Γ(n/r) / Γ(l/p + m/q + n/r + 1)

Special case check. For l = m = n = p = q = r = 1 and a = b = c = 1 (the tetrahedron x + y + z ≤ 1): I = 1/Γ(4) = 1/6, which is the correct volume.