ON THE EVALUATION OF MULTIPLE INTEGRALS

K. L. D. Gunawardena

University of Wisconsin – Oshkosh

1. Introduction. The students in my Mathematical Statistics course came across the following double integrals, and were unable to evaluate them using techniques that they had learned in calculus courses. The two integrals were

∫_{0}^{1}∫_{0}^{1} \frac{θ^{2}(x_{1}x_{2})^{θ−1}}{ln(x_{1}x_{2})} dx_{1} dx_{2} and ∫_{0}^{∞}∫_{0}^{∞} \frac{θ^{2}[(1 + x_{1})(1 + x_{2})]^{−(θ+1)}}{ln[(1 + x_{1})(1 + x_{2})]} dx_{1} dx_{2}

where θ > 0. The computer algebra systems Derive and Mathematica were unable to evaluate the integrals. However, using results from Probability Theory the two integrals can be evaluated. We will generalize the two double integrals and instead consider the two multiple integrals

∫_{0}^{1} ⋯ ∫_{0}^{1} \frac{θ^{n}(x_{1} ⋯ x_{n})^{θ−1}}{ln(x_{1} ⋯ x_{n})} dx_{1} ⋯ dx_{n}

and

∫_{0}^{∞} ⋯ ∫_{0}^{∞} \frac{θ^{n}[(1 + x_{1}) ⋯ (1 + x_{n})]^{−(θ+1)}}{ln[(1 + x_{1}) ⋯ (1 + x_{n})]} dx_{1} ⋯ dx_{n}

where θ > 0 and n is an integer greater than or equal to 2.

The key probability result needed to evaluate the integrals is the following.

Let X_{1}, …, X_{n} be n random variables with joint probability density function (pdf) f(x_{1}, …, x_{n}) and Y = g(X_{1}, …, X_{n}). The expected value E(Y) can be obtained in one of two ways. We could directly calculate

Eg(X_{1}, …, X_{n}) = ∫_{−∞}^{∞} ⋯ ∫_{−∞}^{∞} g(x_{1}, …, x_{n})f(x_{1}, …, x_{n})dx_{1} ⋯ dx_{n}

or we could also find the pdf f_{Y} (y) of Y = g(X_{1}, …, X_{n}) and

Eg(X_{1}, …, X_{n}) = E(Y) = ∫_{−∞}^{∞} yf_{Y} (y)dy.

2. Evaluation of ∫ _{0}^{1} ⋯ ∫_{0}^{1} \frac{θ^{n}(x_{1}···x_{n})^{θ−1}}{ln(x_{1}···x_{n})} dx_{1} ⋯ dx_{n}.

Let X_{1}, …, X_{n} be a random sample of size n from a distribution with pdf f(x) = θx^{θ−1} if 0 < x < 1, and zero otherwise; θ > 0.

Define the random variable Y = (∑_{i=1}^{n} ln X_{i})^{−1}.
Since X_{1}, …, X_{n} is a random sample, the joint pdf of X_{1}, …, X_{n} is

f(x_{1}, …, x_{n}) = ∏_{i=1}^{n} f(x_{i}) = θ^{n}(x_{1} ⋯ x_{n})^{θ−1}, 0 < x_{i} < 1

and

\begin{aligned}E(∑_{i=1}^{n} ln X_{i})^{−1} & = ∫_{0}^{1} ⋯ ∫_{0}^{1}(∑_{i=1}^{n} ln x_{i})^{−1}θ^{n}(x_{1} ⋯ x_{n})^{θ−1}dx_{1} ⋯ dx_{n} \\ & = ∫_{0}^{1} ⋯ ∫_{0}^{1} \frac{θ^{n}(x_{1} ⋯ x_{n})^{θ−1}}{ln(x_{1} ⋯ x_{n})} dx_{1} ⋯ dx_{n}. \end{aligned}

If the random variable X has pdf f(x) = θx^{θ−1} when 0 < x < 1, and zero otherwise, where θ > 0, then the random variable W = − ln X has an exponential distribution with mean 1/θ, since the pdf of W, f_{W} (w) is given by

f_{W} (w) = f(e^{−w})|\frac{d}{dw} e^{−w}| = θ(e^{−w})^{θ−1}| − e^{−w}| = θe^{−θw}

for w > 0.

Since X_{i}, i = 1, …, n is a random sample W_{i} = − ln X_{i} are independent. Also, since W_{i}, i = 1, …, n are independent exponential random variables with mean 1/θ, ∑_{i=1}^{n} W_{i} has a gamma distribution with parameters 1/θ and n. Thus, the random variable U = − ∑_{i=1}^{n} ln X_{i} has pdf

f_{U}(u) = \frac{θ^{n}u^{n−1}e^{−θu}}{Γ(n)} u ≥ 0, and zero otherwise,

and

\begin{aligned}E(Y) & = E(−1/U) = ∫_{0}^{∞} −\frac{1}{u} \frac{θ^{n}u^{n−1}e^{−θu}}{Γ(n)} du \\ & = −\frac{1}{Γ(n)} ∫_{0}^{∞} θ^{n}u^{n−2}e^{−θu}du = −\frac{θΓ(n − 1)}{Γ(n)} = −\frac{θ}{n − 1}. \end{aligned}

Thus,

∫_{0}^{1} ⋯ ∫_{0}^{1} \frac{θ^{n}(x_{1} ⋯ x_{n})^{θ−1}}{ln(x_{1} ⋯ x_{n})} dx_{1} ⋯ dx_{n} = −\frac{θ}{n − 1}.

3. Evaluation of ∫ _{0}^{∞} ⋯ ∫_{0}^{∞} \frac{θ^{n}[(1+x_{1})···(1+x_{n})]^{−(θ+1)}}{ln[(1+x_{1})···(1+x_{n})]} dx_{1} ⋯ dx_{n}.

Let X_{1}, …, X_{n} be a random sample of size n from a distribution with pdf f(x) = θ(1 + x)^{−(θ+1)} if x > 0, and zero otherwise, where θ > 0.

Define the random variable

Y = (∑_{i=1}^{n} ln(1 + X_{i}))^{−1}.

Since X_{1}, …, X_{n} is a random sample, the joint pdf of X_{1}, …, X_{n} is

f(x_{1}, …, x_{n}) = ∏_{i=1}^{n} f(x_{i}) = θ^{n}[(1 + x_{1}) ⋯ (1 + x_{n})]^{−(θ+1)}, x_{i} > 0

and

E(∑_{i=1}^{n} ln(1 + X_{i}))^{−1}

= ∫_{0}^{∞} ⋯ ∫_{0}^{∞}(∑_{i=1}^{n} ln(1 + x_{i}))^{−1}θ^{n}[(1 + x_{1}) ⋯ (1 + x_{n})]^{−(θ+1)}dx_{1} ⋯ dx_{n}

= ∫_{0}^{∞} ⋯ ∫_{0}^{∞} \frac{θ^{n}[(1 + x_{1}) ⋯ (1 + x_{n})]^{−(θ+1)}}{ln[(1 + x_{1}) ⋯ (1 + x_{n})]} dx_{1} ⋯ dx_{n}.

If the random variable X has pdf f(x) = θ(1+x)^{−(θ+1)} when x > 0, and zero otherwise, where θ > 0, then the random variable W = ln(1 + X) has an exponential distribution with mean 1/θ, since the pdf of W, f_{W} (w) is given by

f_{W} (w) = f(e^{w} − 1)|\frac{d}{dw} (e^{w} − 1)| = θe^{−(θ+1)w}|e^{w}| = θe^{−θw} for w > 0.

As before

U = ∑_{i=1}^{n} ln(1 + X_{i})

has a gamma distribution with parameters 1/θ and n, and

\begin{aligned}E(Y) & = E(1/U) = ∫_{0}^{∞} \frac{1}{u} \frac{θ^{n}u^{n−1}e^{−θu}}{Γ(n)} du \\ & = \frac{1}{Γ(n)} ∫_{0}^{∞} θ^{n}u^{n−2}e^{−θu}du = \frac{θΓ(n − 1)}{Γ(n)} = \frac{θ}{n − 1}. \end{aligned}

Thus,

∫_{0}^{∞} ⋯ ∫_{0}^{∞} \frac{θ^{n}[(1 + x_{1}) ⋯ (1 + x_{n})]^{−(θ+1)}}{ln[(1 + x_{1}) ⋯ (1 + x_{n})]} dx_{1} ⋯ dx_{n} = \frac{θ}{n − 1}.

These integrals were encountered in the derivation of unbiased estimates for the parameter θ of the two pfds considered. The probability results used can be found in any standard text on Mathematical Statistics [1]. It would be of interest to know whether these integrals can be evaluated without using results from Probability Theory.

\underline{Reference}

1. L. J. Bain and M. Engelhardt, Introduction to Probability and Mathematical Statistics, 2nd ed., PWS–Kent, Boston, MA, 1992.