An infinite number of Multiple Integrals

Calculus Level 2

Compute:

∫ 0 1 x 1 d x 1 \displaystyle \int_{0}^{1} x_1 d x_1 + ∫ 0 1 ∫ 0 1 x 1 ∗ x 2 d x 2 d x 1 \displaystyle \int_{0}^{1} \int_{0}^{1} x_1*x_2 \,d x_2\, d x_1 + ∫ 0 1 ∫ 0 1 ∫ 0 1 x 1 ∗ x 2 ∗ x 3 d x 3 d x 2 d x 1 \displaystyle \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} x_1*x_2*x_3 \,d x_3\, d x_2\, d x_1 +

… \ldots + ∫ 0 1 ∫ 0 1 … ∫ 0 1 ∫ 0 1 x 1 ∗ x 2 ∗ … ∗ x n − 1 ∗ x n d x n d x n − 1 … d x 2 d x 1 \displaystyle \int_{0}^{1} \int_{0}^{1} \ldots \int_{0}^{1} \int_{0}^{1} x_1*x_2*\ldots*x_{n-1}*x_n \, d x_n\, d x_{n-1}\, \ldots\, d x_2\, d x_1

as n → ∞ n\rightarrow \infty

The essence of the question is to calculate the sum of all multiple integrals with the bounds 0 and 1 and an inner term of ∏ n = 1 n = ∞ n \displaystyle \prod_{n=1}^{n=\infty} n , starting with ∫ 0 1 x 1 d x 1 \displaystyle \int_{0}^{1} x_1 d x_1 and ending with the infinite multiple integral ∫ 0 1 ∫ 0 1 … ∫ 0 1 ∫ 0 1 x 1 ∗ x 2 ∗ … ∗ x n − 1 ∗ x n d x n d x n − 1 … d x 2 d x 1 \displaystyle \int_{0}^{1} \int_{0}^{1} \ldots \int_{0}^{1} \int_{0}^{1} x_1*x_2*\ldots*x_{n-1}*x_n \, d x_n\, d x_{n-1}\, \ldots\, d x_2\, d x_1 .


The answer is 1.

Peter Burbery by Peter Burbery , 19, USA

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1 solution

Peter Burbery
Apr 10, 2021

First, let's compute the base case integral where only one dimension is present.

∫ 0 1 x 1 d x 1 = 1 2 \displaystyle \int_{0}^{1} x_1 d x_1=\frac{1}{2}

Let's continue computing multiple integrals and see if we notice a pattern:

∫ 0 1 ∫ 0 1 x 1 ∗ x 2 d x 2 d x 1 = 1 4 \displaystyle \int_{0}^{1} \int_{0}^{1} x_1*x_2 \,d x_2\, d x_1=\frac{1}{4}

∫ 0 1 ∫ 0 1 ∫ 0 1 x 1 ∗ x 2 ∗ x 3 d x 3 d x 2 d x 1 = 1 8 \displaystyle \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} x_1*x_2*x_3 \,d x_3\, d x_2\, d x_1=\frac{1}{8}

We now notice that we have the following first terms 1 2 , 1 4 , 1 8 \displaystyle {\frac{1}{2},\frac{1}{4},\frac{1}{8}}

We see a pattern of a n = 1 2 n \displaystyle a_n=\frac{1}{2^n}

The sum of this geometric series is famously known to be 1:

∑ n = 1 n = ∞ 1 2 n = 1 2 1 − 1 2 = 1 2 1 2 = 1 \displaystyle \sum_{n=1}^{n=\infty} {\frac{1}{2^n}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1

And so we arrive at 1 as our answer.

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