Question

Many improper double integrals may be handled using the techniques for improper integrals in one variable (Section 7.8 ). For example, under suitable conditions on \(f\), \(\int_{a}^{\infty} \int_{g(x)}^{h(x)} f(x, y) d y d x=\lim _{b \rightarrow \infty} \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) d y d x\). Use or extend the one-variable methods for improper integrals to evaluate the following integrals. $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)} d y d x$$

Short answer

Question: Calculate the value of the improper double integral $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)} d y d x$$ Answer: The value of the improper double integral is $$\pi^2$$.

Step-by-step solution

Convert the double integral into a limit expression

Since the integral is over the whole x-y plane, we will convert the double integral to the proper form by taking limits. The proper form is given as: $$\lim_{a\to-\infty} \lim_{b\to\infty} \lim_{c\to-\infty} \lim_{d\to\infty} \int_{a}^{b} \int_{c}^{d} \frac{1}{(x^2+1)(y^2+1)}dydx$$

Integrate the inner integral with respect to y

Now, we will integrate the inner integral while keeping the limits of the outer integral unchanged: $$\lim_{a\to-\infty} \lim_{b\to\infty} \lim_{c\to-\infty} \lim_{d\to\infty} \int_{a}^{b} \left[\frac{\arctan(y)}{(x^2+1)}\right]_{c}^{d} dx$$

Evaluate the result of the inner integral with limits

Substitute the limits into the expression we got in the previous step: $$\lim_{a\to-\infty} \lim_{b\to\infty} \lim_{c\to-\infty} \lim_{d\to\infty} \int_{a}^{b} \frac{\arctan(d)-\arctan(c)}{(x^2+1)} dx$$ As \(\lim_{d\to\infty} \arctan (d) = \frac{\pi}{2}\) and \(\lim_{c\to-\infty} \arctan (c) = -\frac{\pi}{2}\), the expression becomes: $$\lim_{a\to-\infty} \lim_{b\to\infty} \int_{a}^{b} \frac{\pi}{(x^2+1)} dx$$

Integrate the outer integral with respect to x

Now, we will integrate the outer integral: $$\lim_{a\to-\infty} \lim_{b\to\infty} \left[\pi \arctan(x)\right]_{a}^{b}$$

Evaluate the result of the outer integral with limits

Substitute the limits into the expression we got in the previous step: $$\pi \left(\lim_{b\to\infty} \arctan(b) - \lim_{a\to-\infty} \arctan(a)\right)$$ As \(\lim_{b\to\infty} \arctan(b) = \frac{\pi}{2}\) and \(\lim_{a\to-\infty} \arctan(a) = -\frac{\pi}{2}\), the expression becomes: $$\pi\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)$$

Simplify the expression to get the final answer

Solving the expression, we get the final answer: $$\pi\left(\frac{\pi}{2}+\frac{\pi}{2}\right)=\pi\left(\pi\right)=\pi^2$$ So, the value of the improper double integral is: $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)} d y d x=\pi^2$$

Key concepts

Limits

Limits are a crucial concept in calculus used to understand the behavior of a function as its input approaches a certain value or infinity. When dealing with double integrals over infinite regions, limits help us transition from an improper integral to a proper one, making it easier to evaluate. In the context of the provided exercise, we encounter limits when converting our problem of integrating over the entire plane:

• The use of limits transforms the improper double integral into a form that can be handled systematically.
• By introducing limits as variables tend to plus or minus infinity, we define the region of integration.
• We systematically evaluate the integrals from these finite bounds and analyze their behavior as they approach infinite boundaries.

For example, in the exercise, we transformed our integral by imagining boundaries at negative infinity and infinity, and used limits to approximate its bounds and solve it step by step, thus allowing the definition of the proper integral.

Arctangent

The arctangent function, denoted as \( ext{arctan}(x)\), inversely relates to the tangent function. It is essential when integrating functions that arise from circular or trigonometric identities. In the solution provided, we see \( ext{arctan}(y)\) occurring as a result of the integration with respect to \(y\).

• The inner integral involves an expression \(\frac{1}{y^2+1}\), which, when integrated, yields \( ext{arctan}(y)\). This is a standard result since the derivative of \( ext{arctan}(y)\) is \(\frac{1}{y^2+1}\).
• The limits \(\lim_{d \to \infty} ext{arctan}(d) = \frac{\pi}{2}\) and \(\lim_{c \to -\infty} ext{arctan}(c) = -\frac{\pi}{2}\) imply the asymptotic behavior of the function due to its horizontal asymptotes.

Arctangent functions appear often when using integration techniques involving trigonometric substitutions, particularly for areas under curves related to secant and tangent.

Integration Techniques

Integration techniques are strategies used to evaluate integrals, especially those that are not straightforward. In the exercise, several techniques come together to address the improper double integral.One key technique is transitioning improper integrals into a solvable format using limits, making them proper. Then we directly integrate the simpler functions we derive. Here is how the technique is applied in the problem:

• **Inner integration**: We first integrate with respect to \(y\) using knowledge of the \(\frac{1}{y^2+1}\) integrand, resulting in \(\text{arctan}(y)\).
• **Outer integration**: The result is further simplified to another integral with respect to \(x\), given by \(\frac{\pi}{(x^2+1)}\), which mirrors the inner integration technique.
• **Substitution of limits**: Following the integration, we substitute infinity limits back to evaluate the integral completely, using the properties of \( ext{arctan}\) and its limits.

These techniques showcase the elegance of calculus in breaking down complex double integrals into manageable single-variable problems and systematically approaching them through structured limits and integration.