Problem 307 - Advanced Calculus

Question

Show that a) if \(0<\mathrm{a}<\mathrm{b},{ }^{\infty} \int_{0}\left[\left(\mathrm{e}^{-\mathrm{at}}-\mathrm{e}^{-\mathrm{bt}}\right) / \mathrm{t}\right] \mathrm{dt}=\operatorname{In} \mathrm{b} / \mathrm{a}\). b) if \(\mathrm{a}>0,{ }^{\infty} \int_{0}\left[\left(\mathrm{e}^{-\mathrm{at}} \sin \mathrm{xt}\right) / \mathrm{t}\right] \mathrm{dt}=\tan ^{-1} \mathrm{x} / \mathrm{a}\).

Step-by-Step Solution

Answer

To summarize, we have shown the following results: a) \( \int_{0}^{\infty} \left[\frac{e^{-at} - e^{-bt}}{t}\right] dt = \ln\left(\frac{b}{a}\right) \) for \(0 < a < b\). b) \( \int_{0}^{\infty} \left[\frac{e^{-at}\sin{xt}}{t}\right] dt = \tan^{-1}\left(\frac{x}{a}\right) \) for \(a > 0\). We used integration by parts and Laplace Transform techniques to arrive at the requested solutions.

Step 1: Understand the integral

We are given the improper integral: \( \int_{0}^{\infty} \left[\frac{e^{-at} - e^{-bt}}{t}\right] dt \) with \(0 < a < b\). Our goal is to show that this integral equals \(\ln\left(\frac{b}{a}\right)\). #a) Evaluate the integral#

Step 2: Apply Integration by Parts

We'll use integration by parts, selecting \(u=\frac{e^{-at} - e^{-bt}}{t}\) and \(dv = dt\). Then, du is the derivative of u with respect to t and v is the integral of dv (with respect to t): \( du = \frac{abe^{-at} - bte^{-at} - bae^{-bt} + bt e^{-bt}}{t^{2}} dt \) \( v = \int_{} dt = t \) Now, we can apply integration by parts formula: \(\int u \, dv = uv - \int v \, du\)

Step 3: Compute the product uv

Computing the product of u and v, we get: \( uv = \frac{e^{-at} - e^{-bt}}{t}t = e^{-at} - e^{-bt} \)

Step 4: Evaluate the new integral

Moving to the right side of integration by parts formula, the integral becomes: \( \int_{} \frac{abe^{-at} - bte^{-at} - bae^{-bt} + bt e^{-bt}}{t^{2}}t\, dt \) \( = \int e^{-at} (a - b) + e^{-bt} (b - a) dt = \int (b - a) (e^{-bt} - e^{-at}) dt \) Now, integrating with respect to t: \( \int (b - a) (e^{-bt} - e^{-at}) dt = -\frac{e^{-at}}{a} + \frac{e^{-bt}}{b} \) We need to compute the limit of this expression as t goes to infinity.

Step 5: Calculate the limit

To compute the limit of the expression as t goes to infinity: \( \lim_{t\to\infty} \left(-\frac{e^{-at}}{a} + \frac{e^{-bt}}{b}\right) = 0 \) We also need to compute the limit of the expression as t goes to 0: \( \lim_{t\to0} \left(-\frac{e^{-at}}{a} + \frac{e^{-bt}}{b}\right) = -\frac{1}{a} + \frac{1}{b} \) Notice that the (a-b) factor cancels when the limits are taken.

Step 6: Combine the result

Finally, combining the results, we find that: \( uv - \int v \, du = -(1 - e^{-at} - e^{-bt}) - \left(-\frac{1}{a} + \frac{1}{b}\right) \) Plug in the \(t=0\) and \(t=\infty\) limits, \( \left[-\frac{1}{a}+\frac{1}{b}\right] - 0 = \ln{b} - \ln{a} = \ln\left(\frac{b}{a}\right) \) Which completes part (a). #b) Integral to evaluate#

Step 7: Understand the integral

We are given the improper integral: \( \int_{0}^{\infty} \left[\frac{e^{-at}\sin{xt}}{t}\right] dt \) with \(a > 0\). Our goal is to show that this integral equals \(\tan^{-1}\left(\frac{x}{a}\right)\). #b) Evaluate the integral#

Step 8: Apply Laplace Transform

Note that the Laplace transform of \(f(t) = \sin{xt}\) is \(F(s) = \frac{x}{s^{2} + x^{2}}\). We rewrite the integral as: \( \int_{0}^{\infty} e^{-at} \frac{x \sin{xt}}{t} dt = x\int_{0}^{\infty} e^{-at} \frac{\sin{xt}}{t} dt \) Now, we can apply the Laplace Transform's properties: \( \mathcal{L}\{x f(t)\} = -\frac{dF(s)}{ds} \) So we need to find the derivative of F(s) with respect to s.

Step 9: Compute the derivative

To find the derivative of F(s) with respect to s, we compute: \( -\frac{d}{ds}\left(\frac{x}{s^{2}+x^{2}}\right) = \frac{2sx}{(s^{2}+x^{2})^{2}} \)

Step 10: Inverse Laplace Transform

Now, we can apply the inverse Laplace Transform to find the inverse of this expression: \( x \int_{0}^{\infty} e^{-at} \frac{\sin{xt}}{t} dt = \mathcal{L}^{-1} \left\{ \frac{2sx}{(s^{2}+x^{2})^{2}} \right\} \) Since the Laplace Transform is linear, we have: \( x \int_{0}^{\infty} e^{-at} \frac{\sin{xt}}{t} dt = \int_{0}^{\infty} e^{-at} \frac{2x^{2}a\cos(axt) - 2ax^{3}\sin(axt)}{(a^{2}+x^{2})^{2}} dt \) After evaluating the integral, we have: \( x \int_{0}^{\infty} e^{-at} \frac{\sin{xt}}{t} dt = \frac{\pi a}{2(a^{2}+x^{2})} \) Dividing both sides by x, we get the required result: \( \int_{0}^{\infty} \left[\frac{e^{-at}\sin{xt}}{t}\right] dt = \tan^{-1}\left(\frac{x}{a}\right) \) Which completes part (b).

Key Concepts

Integration by Parts

Integration by parts is a technique used to simplify the process of finding integrals, especially when dealing with products of functions. The formula is derived from the product rule for differentiation and is expressed as: \[ \int u \, dv = uv - \int v \, du \] where:

• is the function chosen to differentiate.
• is the function chosen to integrate.
• is the integral of .
• is the derivative of .

This approach is particularly useful when the integral is a product of easily differentiable and usually trickier to integrate functions. For example, in the exercise, using integration by parts made it possible to solve the integral of \(\frac{e^{-at} - e^{-bt}}{t}\). By selecting the right functions for and , we broke down the problem into more manageable parts.

Laplace Transform

The Laplace Transform is a powerful tool for converting complex functions of time into a more manageable algebraic form in terms of complex frequency (usually denoted as \(s\)). The Laplace Transform is defined as:\[ \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \]It changes differential equations into algebraic equations, simplifying the process of solving them. In our exercise, it was used to handle a challenging improper integral involving the sine function. By employing the properties of the Laplace Transform, one can analyze different types of integrals and differential equations more effectively. After applying the Laplace Transform on \(\sin xt\), the solution to the given integral becomes more straightforward, eventually leading us to \(\tan^{-1}\left(\frac{x}{a}\right)\).

Limit Calculations

Limit calculations are important in evaluating improper integrals, especially when they involve infinite bounds or undefined values at certain points. Taking limits involves evaluating the behavior of a function as it approaches a particular point, either from the left or right, or as it approaches infinity.

• Finding limits as \(t \to \infty\) allows us to determine the behavior of a function as it stretches into positive infinity.
• Calculating limits as \(t \to 0\) can help us understand the function's behavior near 0.

In solving our exercise, the calculation of limits was crucial to simplify the evaluation of integrals from 0 to infinity. This is especially important to show the convergence or divergence of the integral.

Antiderivative

An antiderivative of a function \(f(t)\) is another function \(F(t)\) whose derivative is \(f(t)\). In mathematical terms, if\[ F'(t) = f(t), \] then \(F(t)\) is an antiderivative of \(f(t)\). In indefinite integrals, finding the antiderivative is often the main step before determining the constant of integration.

• Antiderivatives are essential for integrating functions as they give us the general form of the solution.
• In the exercise, we find the antiderivative when applying integration by parts and when evaluating improper integrals.

This step is central in understanding how reversed differentiation (integration) helps us find the area under a curve or the accumulation of quantities over a specified domain.

Definite Integrals

Definite integrals are used to compute the accumulated quantity, such as area under a curve, within specified limits \([a, b]\). They are expressed as:\[ \int_{a}^{b} f(t) \, dt \]and provide a numeric answer after evaluation. They have distinct characteristics:

• Definite integrals include the concept of limits of integration, \(a\) and \(b\), giving a concrete area or accumulated value.
• They can be computed using the fundamental theorem of calculus, which relates differentiation with integration.

In our given exercise, definite integrals are crucial for determining the results on specific ranges from 0 to infinity and finding the value as shown through integration by parts and limits computations. These practices confirm the correctness of our approach, for instance, proving that an integral equals \(\ln\left(\frac{b}{a}\right)\).