Problem 77 - Essential Calculus

Question

(a) The improper integrals \(\int_{1}^{\infty} \frac{1}{x} d x \quad\) and \(\int_{1}^{\infty} \frac{1}{x^{2}} d x\) diverge and converge, respectively. Describe the essential differences between the integrands that cause one integral to converge and the other to diverge. (b) Sketch a graph of the function \(y=\sin x / x\) over the interval \((1, \infty)\). Use your knowledge of the definite integral to make an inference as to whether or not the integral \(\int_{1}^{\infty} \frac{\sin x}{x} d x\) converges. Give reasons for your answer. (c) Use one iteration of integration by parts on the integral in part (b) to determine its divergence or convergence.

Step-by-Step Solution

Answer

First integral \(\int_{1}^\infty \frac{1}{x} dx\) diverges. Second integral \(\int_{1}^\infty \frac{1}{x^{2}} dx\) converges. The function \(y=\sin x / x\) over the interval \((1, \infty)\) diminishes as \(x\) gets larger, predicting that the integral \(\int_{1}^\infty \frac{\sin x}{x} dx\) also converges. Integration by parts confirmed this intuition, as the integration resulted in the sum of a definite integral from 1 to ∞ of a function smaller than \(\frac{1}{x^2}\) which we know converges, and thus the original integral also converges.

Step 1: Evaluate First Two Integrals

Evaluate the improper integrals\(\int_{1}^{\infty} \frac{1}{x} d x \) and \(\int_{1}^{\infty} \frac{1}{x^{2}} d x\). Using standard integration, the first integral evaluates to \([ \ln |x| ]_{1}^{\infty}\) and after evaluating the limits results in \( \infty - \ln 1\) breaking the limit rule, i.e., infinity minus a finite number, hence, it diverges. The second integral evaluates to \([- \frac{1}{x} ]_{1}^{\infty}\) and after evaluating the limits it is \( -0 - -1 = 1 \) so it converges.

Step 2: Analyze Integrands

The primary difference between the integrands that is causing one integral to diverge, and the other to converge is the exponent of \(x\) in the denominator. As the exponent increases, the value of the denominator increases significantly causing the whole fraction to approach 0 faster, therefore causing the integral to converge.

Step 3: Sketch Function and Analyze

Sketch the function \(y = \sin x / x\) over the interval \((1, \infty)\) and infer the convergence or divergence of \(\int_{1}^{\infty} \frac{\sin x}{x} d x\). For \(x > 1\), \(\sin x / x\) oscillates between positive and negative values. As \(x\) gets larger, the amplitude of oscillations diminishes due to the division by \(x\), suggesting the integral of the function could converge.

Step 4: Integrate By Parts

To prove the inference in Step 3, we can use integration by parts on the integral \(\int_{1}^{\infty} \frac{\sin x}{x} d x\). The integration by parts formula is \(\int udv = uv - \int vdu\). Let \(u=\frac{1}{x}\) and \(dv=\sin x\ dx\). Then \(du=-\frac{1}{x^2}dx\) and \(v=-\cos x\). Using these values in the integration by parts formula, we get \( \left[-\frac{\cos x}{x}\right]_1^\infty + \int_1^\infty \frac{\cos x}{x^2} dx \). The first term evaluates to 0 because both terms within the brackets tend to 0. The second term is an integral of a function smaller than \(\frac{1}{x^2}\) over [1, ∞) which, as we know from before, converges. Therefore, the integral in question also converges.

Key Concepts

Convergence

Convergence is a key concept when evaluating improper integrals. In simple terms, an integral converges if the area under the curve of the function being integrated approaches a finite limit as it extends to infinity. In the context of the exercise, \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges because as \( x \) increases, the integrand \( \frac{1}{x^2} \) becomes significantly small, leading the overall area to reach a finite value.
To identify convergence, consider:

• The behavior of the function as \( x \) approaches infinity.
• If the "tail" of the function is diminishing quickly enough to result in a finite area.

Recognizing when an integral converges is crucial in calculus as it helps us understand the behavior of functions over unbounded intervals.

Divergence

An improper integral diverges if the area under the curve becomes infinite as \( x \) approaches a certain limit or infinity. This means that the integral does not settle into a finite value. Consider the integral \( \int_{1}^{\infty} \frac{1}{x} \, dx \), which diverges.
The key reason behind the divergence here is due to the nature of the integrand. Since \( \frac{1}{x} \) decreases more slowly compared to functions with higher powers in the denominator, its area does not limit to a finite number.

• Evaluate the integrand's rate of decrease.
• If it decreases too slowly, the total area under the curve extends to infinity.

Understanding divergence helps us realize why some functions cannot yield a finite result when integrated over an infinite range.

Integration by Parts

Integration by parts is a technique used to simplify integrals where standard methods might not readily apply. In our exercise, integration by parts was applied to \( \int_{1}^{\infty} \frac{\sin x}{x} \, dx \). This method helps in breaking the integral into simpler parts.
To use integration by parts, remember the formula: \[ \int u\,dv = uv - \int v\,du \] When choosing \( u \) and \( dv \):

• Pick \( u \) as the part of the integrand that simplifies when differentiated.
• Choose \( dv \) as what is left, which should become easier upon integration.

In our example, choosing \( u = \frac{1}{x} \) and \( dv = \sin x \, dx \) led us to a simpler form, confirming the convergence of our integral.
Integration by parts is a versatile tool, making complex integrals much more manageable by strategically simplifying components.

Trigonometric Integrals

Trigonometric integrals involve the integration of functions with trigonometric terms. In the exercise, we dealt with the integral \( \int_{1}^{\infty} \frac{\sin x}{x} \, dx \). Such integrals often require special techniques due to the oscillatory nature of sine and cosine functions.
An interesting point with these integrals is how they behave at infinity. The function \( \sin x / x \) oscillates, but due to the division by \( x \), it diminishes in magnitude as \( x \) increases.

• Graphically, observe how the oscillations become less significant.
• Mathematically, integration by parts or comparison to other known integrals helps determine convergence or divergence.

Understanding trigonometric integrals is crucial as they frequently appear in physics and engineering, offering insight into wave-like behavior and signal processing.