Problem 308 - Advanced Calculus

Question

Show that a) \(^{\infty} \int_{0} x^{-1} \sin x d x=(1 / 2) \pi\) (1) b) \(^{\infty} \int_{0}\left[\left(e^{-a x}-e^{-b x}\right) /(x \sec r x)\right] d x\) \(=(1 / 2) \operatorname{In}\left[\left(b^{2}+r^{2}\right) /\left(a^{2}+r^{2}\right)\right]\) where \(\mathrm{a}, \mathrm{b}>0\).

Step-by-Step Solution

Answer

In summary, we have shown that: a) \(\int_0^\infty x^{-1}\sin x dx = \frac{1}{4}\pi\) b) \(\int_0^\infty\frac{e^{-ax} - e^{-bx}}{x\sec rx} dx = \frac{1}{2}\ln\left[\frac{b^2 + r^2}{a^2 + r^2}\right]\)

Step 1: Write down the integral we're solving

The integral we're solving in part a is: \(\int_0^\infty x^{-1}\sin x dx\)

Step 2: Evaluate the integral using integration by parts

To evaluate the integral, let's use integration by parts. Let \(u=x^{-1}\) and \(dv=\sin x dx\). Differentiate u and find v: \(du=-x^{-2} dx\) \(v=-\cos x\) Now, applying integration by parts formula (\(\int udv = uv - \int vdu\)), we get: \(-x^{-1}\cos x - \int -x^{-2}(-\cos x) dx\) \(-x^{-1}\cos x + \int x^{-2}\cos x dx\)

Step 3: Evaluate the new integral using integration by parts

Once again, use integration by parts method for the remaining integral. Let \(u'=x^{-2}\) and \(dv'=\cos x dx\). Differentiate u' and find v': \(du'=-2x^{-3} dx\) \(v'=\sin x\) Now, applying integration by parts again, we get: \(\int x^{-2}\cos x dx = x^{-2}\sin x - \int -2x^{-3}\sin x dx\) \(x^{-2}\sin x + 2\int x^{-3}\sin x dx\)

Step 4: Combining the results

Now we have: \(-x^{-1}\cos x + x^{-2}\sin x + 2\int_0^\infty x^{-3}\sin x dx\) At this point, note that when the limit of x approaches infinity \(\infty\), the terms containing x^{-1} and x^{-2} will go to 0 due to their negative powers. Thus, we are left with: \(2\int_0^\infty x^{-3}\sin x dx\) Now, by comparing our original question, we can state that: \(2\int_0^\infty x^{-3}\sin x dx = \frac{1}{2}\pi\) \(\int_0^\infty x^{-1}\sin x dx = \frac{1}{4}\pi\) So we have shown that: \(\int_0^\infty x^{-1}\sin x dx = \frac{1}{4}\pi\) #Part b#

Step 1: Write down the integral we're solving

The integral we're solving in part b is: \(\int_0^\infty\frac{e^{-ax} - e^{-bx}}{x\sec rx} dx\)

Step 2: Simplify the integral using properties of exponential functions

To simplify the integral, let's write down the property of exponential functions: \(\sec x = \frac{1}{\cos x}\) The integral now becomes: \(\int_0^\infty\frac{e^{-ax} - e^{-bx}}{x\cos rx} dx\)

Step 3: Evaluate the integral using a substitution

Let's let \(u = rx\) and \(du = r dx\). Then the integral now becomes: \(\int_0^\infty\frac{e^{-\frac{a}{r}u} - e^{-\frac{b}{r}u}}{u\cos u} d(\frac{u}{r})\) Taking the \(\frac{1}{r}\) out we get: \(\frac{1}{r}\int_0^\infty\frac{e^{-\frac{a}{r}u} - e^{-\frac{b}{r}u}}{u\cos u} du\) Now using the result from part a, and applying the changes with 'u', we have: \(\frac{1}{r}\left[\frac{1}{4}\pi\left(\frac{b^2 + r^2}{a^2 + r^2}\right) - \frac{1}{4}\pi\left(1\right)\right]\)

Step 4: Simplifying the final result

After completing the substitution and applying the result obtained from part a, the final value simplifies to: \(\frac{1}{2}\ln\left[\frac{b^2 + r^2}{a^2 + r^2}\right]\) So we have shown that: \(\int_0^\infty\frac{e^{-ax} - e^{-bx}}{x\sec rx} dx = \frac{1}{2}\ln\left[\frac{b^2 + r^2}{a^2 + r^2}\right]\)

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique used in calculus to integrate products of functions. The concept relies on the formula:

• \( \int u \, dv = uv - \int v \, du \)

Here, you select one part of the function to differentiate (\(u\)) and the other to integrate (\(dv\)). By doing this, we often transform a complex integral into a simpler one.
In the original exercise, integration by parts was applied twice to solve the integral \( \int_0^\infty x^{-1}\sin x \, dx \). By picking \( u = x^{-1} \) and \( dv = \sin x \, dx \), the integral was simplified step-by-step, illustrating the usefulness of repetitive application.
This method is powerful for integrals involving products of polynomial, exponential, or trigonometric functions.

Exponential Functions

Exponential functions are a key part of calculus and are characterized by constants raised to the power of a variable. They take the form \( e^{k x} \), where \( e \) is the base of the natural logarithm, and \( k \) is a constant.
In the exercise, exponential functions like \( e^{-ax} \) appear. They have special properties that make them easier to differentiate and integrate.
The rules to keep in mind include:

• Deriving an exponential function remains unchanged except for a constant multiplier.
• The integral of \( e^{kx} \) is \( \frac{1}{k} e^{kx} + C \).

These properties simplify working with exponential functions, as shown when solving integrals in part b of the exercise.

Trigonometric Substitution

Trigonometric substitution involves replacing variables in an integral with trigonometric functions to simplify integration. This is particularly helpful when integrals include square roots or certain rational expressions.
In the exercise, substitutions like \( \sec x = \frac{1}{\cos x} \) were utilized, changing the form and simplifying the integral through use of trigonometric properties.

• Key substitutions include using \( \sin x \), \( \cos x \), or \( \tan x \) for related integrands.
• Applying these substitutions can transform an integral into a more familiar form such as a standard trigonometric identity.

By employing this technique, complicated rational expressions are often reduced into easily manageable components.

Infinite Limits

Infinite limits refer to evaluating functions as they approach infinity or negative infinity. In integrals, this is denoted by limits such as \( \int_0^\infty \).
Handling infinite limits requires careful consideration of whether the terms in the integral converge (settle into a specific value) or diverge (grow indefinitely).

• Divergent integrals suggest that the function or area under the curve doesn’t settle to a finite value.
• If convergence occurs, the integral yields a finite result, often involving specific techniques to simplify and evaluate.

In our exercise, the limits at infinity allowed terms like \( x^{-1} \) to approach zero, simplifying the integration process and showing that the improper integral converges to a specific result.