Evaluate the triple integral of xyz times e to the power minus (x + y + z), over the region 0 ≤ x ≤ k, 0 ≤ y ≤ k − x, 0 ≤ z ≤ k − x − y.

Step-by-step solution

1. 1

   Recognise the region as a tetrahedron. The nested limits are exactly the conditions

   $x \ge 0, \quad y \ge 0, \quad z \ge 0, \quad x + y + z \le k$

   the simplex with vertices at the origin and $(k,0,0)$, $(0,k,0)$, $(0,0,k)$. Seeing it this way is what lets the exponential, which depends only on $x+y+z$, be pulled out along level surfaces.

2. 2

   Use the fact that the exponential is constant on each slice. Write $u = x + y + z$. On the slice where $x+y+z = u$ the factor $e^{-u}$ is constant, so the integral reduces to a single integral in $u$ once the "amount" of $xyz$ on each slice is known.

3. 3

   Compute the weight of xyz on the sub-simplex. The scaling law gives

   $\int_{x+y+z\le u} xyz\,dV = \frac{u^6}{720}$

   (a homogeneous degree-3 integrand over a 3-dimensional region scales as $u^{3+3} = u^6$; the constant is $\frac{1}{720} = \frac{1!\,1!\,1!}{6!}$ from the Dirichlet integral). Differentiating with respect to $u$ gives the density on the slice, $\frac{u^5}{120}$.

4. 4

   Reduce to a one-dimensional integral.

   $I = \frac{1}{120}\int_0^k u^5 e^{-u}\,du$

   This is an incomplete gamma integral, and repeated integration by parts (or the standard antiderivative) gives

   $\int u^5 e^{-u}\,du = -e^{-u}\left(u^5 + 5u^4 + 20u^3 + 60u^2 + 120u + 120\right)$

5. 5

   Evaluate between 0 and k.

   $\int_0^k u^5 e^{-u}\,du = 120 - e^{-k}\left(k^5 + 5k^4 + 20k^3 + 60k^2 + 120k + 120\right)$

   Dividing by $120$:

   $I = 1 - e^{-k}\left(1 + k + \frac{k^2}{2} + \frac{k^3}{6} + \frac{k^4}{24} + \frac{k^5}{120}\right)$

   The bracket is the first six terms of the Maclaurin series for $e^{k}$ — this is the regularised incomplete gamma function $P(6, k)$, equivalently the probability that a Poisson variable with mean $k$ is at least $6$.

6. 6

   Check numerically and at the limits. As $k \to \infty$, $I \to 1$; as $k \to 0$, $I \to 0$. Direct nested quadrature over the tetrahedron gives $0.0005941848$ at $k = 1$, $0.0420210382$ at $k = 2.5$ and $0.3840393452$ at $k = 5$, matching the closed form to ten decimals in each case.