Problem 11 - Thomas Calculus

Question

Evaluate the integral $$ \int_{0}^{\infty} \frac{e^{-a x}-e^{-b x}}{x} d x $$ (Hint: Use the relation $$ \frac{e^{-a x}-e^{-b x}}{x}=\int_{a}^{b} e^{-x y} d y $$ to form a double integral and evaluate the integral by changing the order of integration.)

Step-by-Step Solution

Answer

The integral evaluates to \( \ln \left( \frac{b}{a} \right) \).

Step 1: Understanding the Given Hint

We are provided with a relation: \( \frac{e^{-ax} - e^{-bx}}{x} = \int_{a}^{b} e^{-xy} \, dy \). This suggests rewriting our integral in terms of another integral, changing our approach from a single integral to a double integral.

Step 2: Setting Up the Double Integral

Start by applying the hint to write the original integral as a double integral:\[\int_{0}^{\infty} \frac{e^{-ax} - e^{-bx}}{x} \, dx = \int_{0}^{\infty} \left( \int_{a}^{b} e^{-xy} \, dy \right) \, dx.\] This is equivalent to \( \int_{a}^{b} \left( \int_{0}^{\infty} e^{-xy} \, dx \right) \, dy \).

Step 3: Evaluating the Inner Integral

The inner integral is \( \int_{0}^{\infty} e^{-xy} \, dx \). Treat \( y \) as constant and evaluate:\[\int_{0}^{\infty} e^{-xy} \, dx = \left[ -\frac{e^{-xy}}{y} \right]_{0}^{\infty} = \frac{1}{y}.\]This uses the fact that \( e^{-xy} \to 0 \) as \( x \to \infty \).

Step 4: Substitute the Inner Integral Result

Substitute \( \int_{0}^{\infty} e^{-xy} \, dx = \frac{1}{y} \) back into the outer integral:\[\int_{a}^{b} \frac{1}{y} \, dy.\]

Step 5: Evaluating the Outer Integral

Now, evaluate the outer integral:\[\int_{a}^{b} \frac{1}{y} \, dy = \left[ \ln |y| \right]_{a}^{b} = \ln |b| - \ln |a| = \ln \left( \frac{b}{a} \right).\]

Step 6: Final Result

The value of the original integral is \( \ln \left( \frac{b}{a} \right) \). This result comes from the manipulation and evaluation of the given form using integration techniques.

Key Concepts

Understanding Double Integrals

A double integral is an extension of a single integral. It allows us to integrate a function over a two-dimensional area. Imagine slicing cheese - each slice represents an integral over one dimension.
When stacked, they cover a whole area. Here’s the process simplified:

• First, integrate with respect to one variable. Treat the other variable as a constant.
• Next, integrate the result with respect to the second variable.

In the exercise, using double integrals helps by providing a way to break down the original integral into manageable parts. This is especially useful in complex problems like this one, where functions of two variables are involved.

Changing Order of Integration

The order of integration refers to the sequence in which integrations are performed within a double integral.
Sometimes, we need to change this order to simplify the calculation process. Changing the order can yield an easier inner integral, facilitating simplification. The decision regarding the order might be influenced by:

• The limits of the integral, making sure they remain valid after the change.
• The nature of the functions involved, affording a more straightforward evaluation.

In the exercise, changing the order allowed the integration over the exponential function to be easier, streamlining the entire evaluation process.

Exploring Improper Integrals

Improper integrals arise when an integral has infinite limits or an integrand with an infinite discontinuity.
This exercise featured improper integrals. To handle these:

• Approach them by taking limits - treat the infinite part as a limit problem.
• Focus on convergence - only well-behaved ones converge to a real number.

In the exercise, as part of evaluating the double integral, we tackled an improper integral by considering limits, ensuring the solution was manageable and finite.

Exploring the Exponential Function in Integration

The exponential function is frequently encountered in calculus, given its growth properties and impact on integrals. It often appears in integrals where it exponentially decays, making it important in simplifying and solving integrals.
In the task, the function present was of the form \( e^{-xy} \). Key characteristics include:

• Maintains continuity and differentiability on the real number line.
• Often simplifies into expressions involving natural logarithms, just like in our final result, \( \ln \left( \frac{b}{a} \right) \).

Exponentials make integrals feasible in many forms, especially with bounds that move towards infinity. This transformation is key in taking advantage of their simplification during integration.