   Chapter 15, Problem 5P

Chapter
Section
Textbook Problem

The double integral ∫ 0 1 ∫ 0 1 1 1 − x y d x   d y is an improper integral and could be defined as the limit of double integrals over the rectangle [0, t] × [0, t] as t → 1−. But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that ∫ 0 1 ∫ 0 1 1 1 − x y d x   d y = ∑ n − 1 ∞ 1 n 2

To determine

To show: 010111xydxdy=n=11n2.

Explanation

Given:

The double integral 010111xydxdy is an improper integral and can be defined as the limit of double integrals over the rectangle [0,t]×[0,t] as t1.

Calculation:

As t1, it is observed that the value of |xy|<1. So, the sum of the geometric series is given by, 11xy=n=0(xy)n except at the point (1,1).

010111xydxdy=0101(n=0(xy)n)dxdy=n=0(0101(xy)ndxdy)=n=0(01yndy01xndx)

Integrate it and apply the limit

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