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The other day, in the honors calculus class at Wohascum College, the instructor asked the students to compute \(\int_0^{\infty} \frac{e^{-x}-e^{-2 x}}{x} d x\). One student split up the integral and made the substitution \(u=2 x\) in the second part, concluding that $$ \int_0^{\infty} \frac{e^{-x}-e^{-2 x}}{x} d x=\int_0^{\infty} \frac{e^{-x}}{x} d x-\int_0^{\infty} \frac{e^{-u}}{u} d u=0 $$ The instructor was not too impressed by this, pointing out that for all positive values of \(x, \frac{e^{-x}-e^{-2 x}}{x}\) is positive, so how could the integral be zero? a. Resolve this paradox. b. Eventually a student gave up and asked Mathematica to compute the integral, and an exact answer appeared on the screen: \(\log [2]\) (which is Mathematica's notation for \(\ln 2\) ). Is this answer correct?

Step by step solution

01

- Split the Integral

Let's first split the given integral: \[ \int_0^{\infty} \frac{e^{-x}-e^{-2x}}{x} dx = \int_0^{\infty} \frac{e^{-x}}{x} dx - \int_0^{\infty} \frac{e^{-2x}}{x} dx \]

02

- Substitution for the Second Integral

Perform the substitution \(u = 2x\) in the second integral. Then, \(du = 2dx\) or \(dx = \frac{du}{2}\). We need to adjust the limits of integration accordingly: When \(x = 0\), \(u = 0\), and when \(x = \infty\), \(u = \infty\). Thus, the second integral becomes: \[ \int_0^{\infty} \frac{e^{-2x}}{x} dx = \int_0^{\infty} \frac{e^{-u}}{u} \cdot \frac{du}{2} = \frac{1}{2} \int_0^{\infty} \frac{e^{-u}}{u} du \]

03

- Combine the Integrals

Now, replace the second integral in the original expression: \[ \int_0^{\infty} \frac{e^{-x}-e^{-2x}}{x} dx = \int_0^{\infty} \frac{e^{-x}}{x} dx - \frac{1}{2} \int_0^{\infty} \frac{e^{-u}}{u} du \]

04

- Recognize the Divergent Integral

Recognize that \( \int_0^{\infty} \frac{e^{-x}}{x} dx \) is known as the Exponential Integral, denoted by others as \( E_1(x) \). For \( x = 0 \) this integral is divergent because it tends to infinity. Therefore, the student's calculation was incorrect because it involved divergent integrals that do not straightforwardly cancel each other out. We have: \[ \int_0^{\infty} \frac{e^{-x}}{x} dx = \infty \] and \[ \frac{1}{2} \int_0^{\infty} \frac{e^{-u}}{u} du = \infty/2 \to \infty \]

05

- Confirm the Correct Answer

From the properties of the given integral, we need to verify the value calculated by Mathematica. The correct value of the integral is: \[ \int_0^{\infty} \frac{e^{-x} - e^{-2x}}{x} dx = \ln 2 \] This can be confirmed with higher mathematics involving asymptotic series or by using advanced computational tools.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

calculus problems

Calculus problems are mathematical challenges that often involve finding derivatives or integrals. In this particular problem, we are dealing with improper integrals. An improper integral is an integral where either the interval of integration is infinite or the integrand becomes infinite within the interval of integration.

Let's look at the given problem: We have the integral \(\begin{equation*} \int_0^{\beginfty} \frac{e^{-x}-e^{-2 x}}{x} d x \end{equation*}\). We can see that the problem has an infinite interval (from 0 to ∞), and therefore it is an improper integral. Hence, it's essential to carefully consider the behavior of the function over the entire interval. Through splitting and substitution, we analyze this integral to find a meaningful solution.

exponential integrals

In the context of calculus, exponential integrals often involve integrals of the form \(\begin{equation*} \int_0^{\beginfty} e^{-x^n} \, dx \end{equation*}\). The problem at hand specifically deals with the exponential integral of functions like \(\frac{e^{-x}}{x}\) and \(\frac{e^{-2x}}{x}\).

When we split the original problem, we get two exponential integrals:

• \(\begin{equation*} \int_0^{\beginfty} \frac{e^{-x}}{x} \, dx \end{equation*}\)
• \(\begin{equation*} \int_0^{\beginfty} \frac{e^{-2x}}{x} \, dx \end{equation*}\), which becomes \(\frac{1}{2} \int_0^{\beginfty} \frac{e^{-u}}{u} \, du \) after substitution.

To solve these integrals accurately, we must understand their behavior as they approach their bounds—usually through limits and comparison to known improper integrals.

integral substitution

Integral substitution is a technique used to simplify an integral by making a substitution that transforms the integrand into a more manageable form. This method can also help in changing the limits of integration to fit the new variable.

In our problem, we made the substitution \(\begin{equation*} u = 2x \end{equation*}\) for the second integral, which aided in simplifying it. Here's the detailed substitution process:

\(\begin{equation*} \int_0^{\beginfty} \frac{e^{-2x}}{x} \, dx = \int_0^{\beginfty} \frac{e^{-u}}{u} \, \frac{du}{2} = \frac{1}{2} \int_0^{\beginfty} \frac{e^{-u}}{u} \, du \end{equation*}\) This step is crucial because it transforms an integral that might seem complicated into one that is easier to recognize and handle, using standard forms of exponential integrals.

divergent integrals

A divergent integral is an integral that does not converge to a finite limit. In this exercise, both individual integrals

• \(\begin{equation*} \int_0^{\beginfty} \frac{e^{-x}}{x} \end{equation*}\)
• \(\frac{1}{2} \int_0^{\beginfty} e^{-u} \frac{1}{u} du\)

are divergent.

It's important to recognize this behavior and understand why combining these divergent integrals does not straightforwardly result in zero. Instead, these types of problems often require special tricks or deeper insights, like recognized series expansions or advanced computational tools, to find a meaningful result. Ultimately, knowing that each integral diverges explains why the instructor was skeptical of the student's initial conclusion and validates the correct answer of \(\begin{equation*} \log(2) \end{equation*}\) as computed by Mathematica.