Evaluating The Double Integral Of (x - Y) / (x + Y)^3

Introduction

In this article, we will delve into the evaluation of the double integral ${ \int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dy \, dx }$. Double integrals are a fundamental concept in multivariable calculus, allowing us to calculate the volume under a surface or the integral of a function over a two-dimensional region. The integral presented here involves a rational function with a somewhat intricate form, requiring careful techniques to solve. This exploration will not only demonstrate the step-by-step process of solving this specific integral but also highlight the nuances and potential pitfalls in evaluating double integrals, such as the order of integration and the behavior of the integrand. Understanding these concepts is crucial for various applications in physics, engineering, and other scientific fields where multivariable calculus plays a key role. This article aims to provide a comprehensive guide, ensuring clarity and accuracy in each step, making it an invaluable resource for students and professionals alike.

Problem Statement

We are tasked with evaluating the double integral: ${ \int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dy \, dx }$. This integral represents the signed volume under the surface defined by the function ${ f(x, y) = \frac{x - y}{(x + y)^3} }$ over the unit square ${ [0, 1] \times [0, 1] }$. The challenge lies in the form of the integrand, which involves a rational function with a cubic denominator. Direct integration might seem daunting, but by employing strategic techniques such as partial fraction decomposition or a suitable substitution, we can simplify the integral and find a solution. The order of integration, whether to integrate with respect to y first or x first, can also influence the complexity of the problem. Therefore, careful consideration is needed to choose the most efficient approach. The final result will provide us with a numerical value, representing the net signed volume over the specified region.

Step-by-Step Solution

Step 1: Integrate with Respect to y

We begin by integrating the inner integral with respect to y, treating x as a constant. This involves evaluating ${ \int_0^1 \frac{x - y}{(x + y)^3} \, dy }$. To tackle this, we can use a substitution method. Let ${ u = x + y }$, so ${ du = dy }$ and ${ y = u - x }$. The limits of integration will change accordingly: when ${ y = 0 }$, ${ u = x }$, and when ${ y = 1 }$, ${ u = x + 1 }$. The integral then becomes:

${ \int_x^{x+1} \frac{x - (u - x)}{u^3} \, du = \int_x^{x+1} \frac{2x - u}{u^3} \, du }$

We can split this into two separate integrals:

${ \int_x^{x+1} \left( \frac{2x}{u^3} - \frac{u}{u^3} \right) \, du = \int_x^{x+1} \left( 2x u^{-3} - u^{-2} \right) \, du }$

Now, we can integrate term by term:

${ \left[ -x u^{-2} + u^{-1} \right]_x^{x+1} = \left( \frac{-x}{(x+1)^2} + \frac{1}{x+1} \right) - \left( \frac{-x}{x^2} + \frac{1}{x} \right) }$

Simplifying this expression, we get:

${ \frac{-x}{(x+1)^2} + \frac{1}{x+1} + \frac{1}{x} - \frac{1}{x} = \frac{-x + (x+1)}{(x+1)^2} = \frac{1}{(x+1)^2} }$

So, the result of the inner integral is ${ \frac{1}{(x+1)^2} }$.

Step 2: Integrate with Respect to x

Next, we integrate the result from the previous step with respect to x over the interval ${ [0, 1] }$. This means we need to evaluate ${ \int_0^1 \frac{1}{(x+1)^2} \, dx }$. This is a straightforward integral that can be solved using a simple substitution or by recognizing it as the derivative of ${ -\frac{1}{x+1} }$.

Let's perform the integration:

${ \int_0^1 (x+1)^{-2} \, dx = \left[ -(x+1)^{-1} \right]_0^1 = \left[ -\frac{1}{x+1} \right]_0^1 }$

Evaluating the limits, we get:

${ -\frac{1}{1+1} - \left( -\frac{1}{0+1} \right) = -\frac{1}{2} + 1 = \frac{1}{2} }$

Thus, the result of the outer integral is ${ \frac{1}{2} }$.

Step 3: Combining the Results and Addressing Potential Issues

Combining the results from Step 1 and Step 2, we find that the value of the double integral ${ \int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dy \, dx }$ is ${ \frac{1}{2} }$. This value represents the signed volume under the surface defined by the integrand over the unit square. However, it's crucial to acknowledge a potential issue in this evaluation: the integrand ${ \frac{x - y}{(x + y)^3} }$ has a singularity at ${ x = -y }$. While this singularity doesn't lie within the region of integration ${ [0, 1] \times [0, 1] }$, the behavior of the integrand near the line ${ x = -y }$ can still influence the result. The convergence of the integral needs careful consideration, and in this case, the integral does converge to ${ \frac{1}{2} }$. If we were to reverse the order of integration, i.e., integrate with respect to x first, we would encounter a similar process but with the roles of x and y interchanged. The final result, after careful evaluation, would be ${ -\frac{1}{2} }$. This discrepancy highlights the importance of the order of integration and the properties of the integrand when dealing with improper integrals or singularities.

Alternate Approach: Reversing the Order of Integration

To further illustrate the nuances of double integrals, let's reverse the order of integration and evaluate ${ \int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dx \, dy }$. This will not only provide an alternative perspective but also reveal how the order of integration can impact the outcome, especially when dealing with functions that are not continuous or have singularities.

Step 1: Integrate with Respect to x

First, we integrate with respect to x, treating y as a constant. This means evaluating ${ \int_0^1 \frac{x - y}{(x + y)^3} \, dx }$. We use a similar substitution as before. Let ${ v = x + y }$, so ${ dv = dx }$ and ${ x = v - y }$. The limits of integration change: when ${ x = 0 }$, ${ v = y }$, and when ${ x = 1 }$, ${ v = 1 + y }$. The integral becomes:

${ \int_y^{1+y} \frac{(v - y) - y}{v^3} \, dv = \int_y^{1+y} \frac{v - 2y}{v^3} \, dv }$

We split this into two integrals:

${ \int_y^{1+y} \left( \frac{v}{v^3} - \frac{2y}{v^3} \right) \, dv = \int_y^{1+y} \left( v^{-2} - 2yv^{-3} \right) \, dv }$

Now, integrate term by term:

${ \left[ -v^{-1} + yv^{-2} \right]_y^{1+y} = \left( \frac{-1}{1+y} + \frac{y}{(1+y)^2} \right) - \left( \frac{-1}{y} + \frac{y}{y^2} \right) }$

Simplifying the expression:

${ \frac{-1}{1+y} + \frac{y}{(1+y)^2} + \frac{1}{y} - \frac{1}{y} = \frac{-1(1+y) + y}{(1+y)^2} = \frac{-1 - y + y}{(1+y)^2} = \frac{-1}{(1+y)^2} }$

Thus, the result of the inner integral is ${ \frac{-1}{(1+y)^2} }$.

Step 2: Integrate with Respect to y

Next, we integrate the result from the previous step with respect to y over the interval ${ [0, 1] }$. This involves evaluating ${ \int_0^1 \frac{-1}{(1+y)^2} \, dy }$.

Performing the integration:

${ \int_0^1 -(1+y)^{-2} \, dy = \left[ (1+y)^{-1} \right]_0^1 = \left[ \frac{1}{1+y} \right]_0^1 }$

Evaluating the limits:

${ \frac{1}{1+1} - \frac{1}{1+0} = \frac{1}{2} - 1 = -\frac{1}{2} }$

Thus, the result of the outer integral is ${ -\frac{1}{2} }$.

Step 3: Comparing the Results

By reversing the order of integration, we obtained a different result: ${ -\frac{1}{2} }$ instead of ${ \frac{1}{2} }$. This discrepancy underscores the importance of the order of integration in double integrals, particularly when dealing with functions that are not continuous or have singularities. The function ${ f(x, y) = \frac{x - y}{(x + y)^3} }$ is not absolutely integrable over the region ${ [0, 1] \times [0, 1] }$, meaning that the integral of its absolute value does not converge. This non-absolute integrability is the reason for the different results when the order of integration is changed. In such cases, the double integral is said to be order-dependent, and the result depends on the order in which the integration is performed. This example illustrates a crucial concept in multivariable calculus and highlights the need for careful consideration when evaluating double integrals.

Conclusion

In conclusion, we have successfully evaluated the double integral ${ \int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dy \, dx }$ using a step-by-step approach, which yielded a result of ${ \frac{1}{2} }$. We also explored an alternative approach by reversing the order of integration, leading to a different result of ${ -\frac{1}{2} }$. This discrepancy emphasizes a critical aspect of double integrals: the order of integration matters, especially when dealing with integrands that are not absolutely integrable or have singularities. The function ${ f(x, y) = \frac{x - y}{(x + y)^3} }$ exemplifies this phenomenon, as it is not absolutely integrable over the region ${ [0, 1] \times [0, 1] }$. The presence of a singularity, even if outside the integration region, can influence the convergence and the final result, making the order of integration a significant factor. This exercise provides valuable insights into the intricacies of multivariable calculus and the evaluation of double integrals, highlighting the need for careful analysis and consideration of the properties of the integrand. Understanding these concepts is essential for accurate and meaningful applications in various scientific and engineering disciplines. This exploration not only enhances our ability to solve complex integrals but also deepens our appreciation for the mathematical subtleties involved in multivariable calculus.

By working through this problem, we have reinforced our understanding of substitution techniques, integration limits, and the significance of the order of integration. The alternative approach of reversing the order served as a powerful reminder of the conditions under which Fubini's Theorem applies and the potential pitfalls when it doesn't. This comprehensive analysis equips us with a more nuanced perspective on evaluating double integrals and reinforces the importance of a thorough understanding of the underlying principles of multivariable calculus.