Question

Let \(E=[0,1] \times[0,1]\). Investigate the existence and equality of \(\int_{E} f d m^{2}\), \(\int_{0}^{1} \int_{0}^{1} f(x, y) d x d y\), and \(\int_{0}^{1} \int_{0}^{1} f(x, y) d y d x\) for the following \(f\). a. \(f(x, y)=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)^{-2}\). b. \(f(x, y)=(1-x y)^{-a}(a>0)\). c. \(f(x, y)=\left(x-\frac{1}{2}\right)^{-3}\) if \(0

Short answer

Part a and b have equal, finite integrals for all three forms; part c does not have a finite integral due to a singularity.

Step-by-step solution

Define the Problem for Part A

We need to determine if \( f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2} \) is integrable over \( E = [0, 1] \times [0, 1] \). Specifically, we seek to compare three integrals: \( \int_{E} f \, dm^2 \), \( \int_{0}^{1} \int_{0}^{1} f(x, y) \, dx \, dy \), and \( \int_{0}^{1} \int_{0}^{1} f(x, y) \, dy \, dx \).

Investigating Full Domain Integration

Let's consider the integrals over \( E \). Due to the symmetry of \( f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2} \), we suspect and confirm with calculations that this function is not singular within \([0, 1] \times [0, 1]\). Calculations show that the full domain integral exists and provides a result close to 0 through symmetrical reasoning.

Fubini's Theorem Applicability

Fubini's Theorem allows us to interchange the order of integration if the integrand is continuous (which it is here). Therefore, both iterated integrals equal the double integral. For \( f \) on \([0,1] \times [0,1]\), continuity implies the ordering doesn't matter.

Solution for Part B

Consider \( f(x, y) = (1 - xy)^{-a} \). The function is continuous over \([0, 1]\times[0, 1]\) for any \( a > 0\) since the denominator never reaches 0. The integrals exist and are finite in any order.

Confirming Existence by Fubini for Part B

Same reasoning with part (a) whereby Fubini's Theorem guarantees the iterated integrals are interchangeable and equal each other, yielding the same result across all versions.

Solution for Part C

\( f(x, y) = \left(x - \frac{1}{2}\right)^{-3} \) when \( 0 < y < |x - \frac{1}{2}| \), otherwise 0. This causes problems with integration as \( x \to \frac{1}{2} \), where the function heads towards \( \infty \). As the integral bounds approach this singularity, it disrupts continuity assumptions necessary for Fubini's Theorem.

Implication for Part C Iterated Integrals

The presence of the singularity means neither iterated nor full domain integrals exist as they approach infinity for those bounds.

Key concepts

Iterated Integrals

Iterated integrals provide a method for integrating functions of two variables by breaking the double integral into two single integrals in succession. In the context of evaluating multivariable functions over a rectangular region, we can use iterated integrals to simplify our calculations. This involves integrating first with respect to one variable, and then with respect to the other.For instance, consider the integrals \( \int_{0}^{1} \int_{0}^{1} f(x, y) \ dx \ dy \) and \( \int_{0}^{1} \int_{0}^{1} f(x, y) \ dy \ dx \). By employing iterated integrals, we achieve the same result as a double integral when Fubini's Theorem is applicable, meaning that we have a continuous function or proper conditions that secure the existence of the integral. When we perform these calculations, we check whether the order of integration affects the result. When conditions are met, the order does not alter the outcome thanks to the versatility provided by iterated integrals.

Double Integration

Double integration is a process that allows us to compute the integral of a function over a two-dimensional area. To execute double integration, we integrate over one variable to collapse the dimension, followed by an integration over the second variable, ultimately capturing the sum of values across the entire region.For problem examples such as the one in our exercise, where \( E = [0,1] \times [0,1] \), double integration captures the behavior of \( f(x, y) \) over the entire square region. By closely examining the function and its properties, such as continuity or existence of singularities, we can determine whether the double integral is well-defined.Using double integration, one must verify that the necessary conditions for integration are met, as seen with our examples. These include ensuring the function does not approach infinity within the integration bounds, maintaining continuity across the region in question.

Singularity in Integration

Singularity in integration refers to points within the domain of an integral where the function may become undefined or approach infinity, causing potential issues in computation. Such singularities can disrupt continuity, making integrals challenging or impossible to evaluate.In some cases, as exemplified by part C of the exercise, the function \( f(x, y) = \left(x - \frac{1}{2}\right)^{-3} \) exhibits a singularity as \( x \rightarrow \frac{1}{2} \). Here, the function approaches infinity within the defined region, violating assumptions necessary for integral calculation. Fubini's Theorem, which requires a well-defined continuous integrand for swapping limits of integration, becomes inapplicable under such conditions.To handle integrations with singularities, special techniques or evaluations may be needed, though often, the presence of a singularity deems the integral undefined or infinite, necessitating an adjustment in the problem approach.