Problem 72 - Thomas Calculus

Question

Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section \(8.8 .\) Evaluate the improper integrals as iterated integrals. \begin{equation}\int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2 y)} d x d y\end{equation}

Step-by-Step Solution

Answer

The improper double integral evaluates to \( \frac{1}{2} \).

Step 1: Evaluate the Inner Integral

The given double integral is \( \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2y)} \ dx \ dy \). We first compute the inner integral with respect to \( x \). This is \( \int_{0}^{\infty} x e^{-(x+2y)} \ dx \). This is an improper integral from 0 to infinity, which can be seen as an exponential integral with respect to \( x \). To evaluate this, you can use integration by parts:Let \( u = x \) and \( dv = e^{-(x+2y)} \ dx \). Then \( du = dx \) and \( v = -e^{-(x+2y)} \).Using integration by parts \( \int u \, dv = uv - \int v \, du \):\[ \int x e^{-(x+2y)} \, dx = -x e^{-(x+2y)} \bigg|_0^{\infty} + \int e^{-(x+2y)} \ dx \].The term \(-x e^{-(x+2y)}\) evaluated at infinity is 0 because the exponential function decreases faster than the polynomial increase of \( x \). At 0, the term is also 0. Therefore:\[ \int e^{-(x+2y)} \ dx = -\frac{1}{2 + y} e^{-(x+2y)} \bigg|_0^{\infty} = \frac{1}{2 + y} \].Thus the inner integral evaluates to \( \frac{1}{(1+2y)^2} \).

Step 2: Evaluate the Outer Integral

Now, use the result from Step 1 to evaluate the outer integral, which becomes \( \int_{0}^{\infty} \frac{1}{(1+2y)^2} \ dy \). We need to compute this improper integral by considering it as the limit: \[ \lim_{b \to \infty} \int_{0}^{b} \frac{1}{(1+2y)^2} \, dy \].To solve the integral \( \int \frac{1}{(1+2y)^2} \ dy \), use the substitution \( u = 1 + 2y \), then \( du = 2 \, dy \), or \( dy = \frac{du}{2} \). The limits for \( u \) change to \( 1 \) and \( \infty \) respectively. Now our integral becomes:\[ \frac{1}{2} \int_{1}^{\infty} u^{-2} \, du \].This simplifies to \( \frac{1}{2} \left[ -u^{-1} \right]_{1}^{\infty} = \frac{1}{2} \left( 0 + 1 \right) = \frac{1}{2} \).

Step 3: Combine Both Results

Since the inner integral evaluates to \( \frac{1}{(1+2y)^2} \) and the outer integral of this result evaluates to \( \frac{1}{2} \), the evaluated improper double integral is:\[ \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2y)} \ dx \, dy = \frac{1}{2} \].

Key Concepts

Integration by parts

Integration by parts is a valuable technique particularly useful when dealing with integrals involving products of functions. It's based on the rule \( \int u \, dv = uv - \int v \, du \), where you choose one part of the function as \( u \) and the other as \( dv \). In the exercise, the aim was to simplify the integral of \( x e^{-(x+2y)} \). Here are the steps involved in detail:

• Choose \( u = x \) and \( dv = e^{-(x+2y)} \, dx \).
• Then calculate \( du = dx \) and integrate \( dv \) to get \( v = -e^{-(x+2y)} \).
• Substitute \( u, dv, du, \) and \( v \) into the integration by parts formula.

The procedure results in two parts to solve:

• A boundary term \( -x e^{-(x+2y)} \bigg|_0^{\infty} \) which turns to zero at both boundaries.
• A simpler integral that remains \( \int e^{-(x+2y)} \, dx \), subsequently yielding a useful result.

Integration by parts here transforms a complex integral into a more manageable form, making further processes straightforward.

Iterated integrals

Iterated integrals refer to a process where a double integral is solved by evaluating one variable at a time. In the initial exercise, the goal was to solve a double integral \( \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2y)} \, dx \, dy \). Here, integration is broken into steps:
First, solve the inner integral \( \int_{0}^{\infty} x e^{-(x+2y)} \, dx \), treating \( y \) as a constant. This involved integration techniques, like those described in the integration by parts, simplifying the integral significantly.
Only then do we compute the outer integral \( \int_{0}^{\infty} \frac{1}{(1+2y)^2} \, dy \). This tiered approach simplifies complex multi-variable integrals into simpler, one-variable integrals handled sequentially. This strategy helps clearly separate problems into smaller, solvable tasks.

Improper integrals

Improper integrals expand the concept of integrals to include limits extending to infinity or integrals with unbounded integrands. While the integral of \( e^{-(x+2y)} \) tends towards zero, convergence must be confirmed.
Using limits, we determine this by turning infinite bounds into finite ones through limit processes:

• Take limits, such as \( \lim_{b \to \infty} \int_{0}^{b} \ldots \, dy \), essentially a workaround to handle infinite bounds.
• This regularization reassures that results are consistently within computational reach, adding rigor to computations.

The step by step solution confirms the double integral converges to a finite value, underscoring the importance of handling infinities carefully to ensure accurate results.

Exponential integrals

Exponential integrals are significant in many fields because of their predictable decay properties and are typically of the form \( \int e^{-ax} \, dx \).
Their consistent behavior makes them a common choice to assess convergence or calculate values across infinite bounds. In the given exercise \( x e^{-(x+2y)} \), the combination of a polynomial term and an exponential meant exploring cases where the exponential swiftly approaches zero.

• Being often easy to differentiate and integrate, exponentials simplify evaluations.
• Decaying fast, they ensure polynomial terms don’t grow unbounded, thus promoting convergence.

This integration type is pivotal in bounding problems with potentially infinite ranges, translating into reliable results - as seen where calculations realistically approached zero outside of critical regions.