Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. $$ \int_{0}^{\pi} \frac{d x}{\cos x-1} $$

Step-by-Step Solution

Answer

The integral diverges because it has singularities at both \( x = 0 \) and \( x = \pi \), leading the integral to approach infinity.

Step 1: Recognize Improper Integral

The integral \( \int_{0}^{\pi} \frac{d x}{\cos x-1} \) is classified as improper because its integrand \( \frac{1}{\cos x - 1} \) becomes undefined at certain points within the integration limits. Specifically, \( \cos x - 1 = 0 \) when \( x = \pi \). This results in a division by zero, which requires us to check the behavior of the integral around these points.

Step 2: Explore Integrand Behavior

Rewrite the integrand: \( \cos x - 1 = -2 \sin^2(\frac{x}{2}) \), so \( \frac{1}{\cos x - 1} = -\frac{1}{2 \sin^2(\frac{x}{2})} \). This shows singularity occurs when \( \sin(\frac{x}{2}) = 0 \), particularly at \( x = 0 \) and \( x = \pi \). Such points require us to evaluate the limit of the integral as \( x \) approaches these problematic points.

Step 3: Apply Limit Process

Divide the integral into parts and apply limit processes to handle the improper nature:\[ \int_{0}^{\pi} \frac{d x}{\cos x-1} = \lim_{a \to 0^+} \int_{a}^{\frac{\pi}{2}} \frac{d x}{\cos x-1} + \lim_{b \to \pi^-} \int_{\frac{\pi}{2}}^{b} \frac{d x}{\cos x-1}. \]

Step 4: Evaluate Limits

Evaluate the limits found in Step 3: - As \( a \to 0^+ \), the integrand approaches a form of \( -\frac{1}{2a^2} \) which tends towards negative infinity, hence diverges. Similarly, the integral behaves the same near \( x = \pi \), thus both limits diverge.

Step 5: Conclusion on Convergence

Since both limits diverge as \( a \to 0^+ \) and \( b \to \pi^- \), the original integral does not converge. Thus, the improper integral diverges.

Key Concepts

Convergence and Divergence

Understanding convergence and divergence is vital when dealing with improper integrals. In the context of improper integrals, we assess if the integral approaches a finite value (convergent) or not (divergent) as we extend limits of integration into points of singularity or infinity. For the given integral \( \int_{0}^{\pi} \frac{d x}{\cos x - 1} \), the integral diverges because as we examine the endpoints, particularly at \( x = 0 \) and \( x = \pi \), the integrand tends toward infinite values. When an integral diverges, it implies the area under the curve is "infinitely large" as it does not settle to a finite number. Recognizing divergence early can save time, as no finite area exists. In practical problems, integrals that diverge often signal something significant, such as a boundary condition or a limitation in a physical model. It also specifies that further mathematical tools or techniques might be necessary for understanding or estimation.

Limit Process in Calculus

A critical aspect of handling improper integrals is the limit process. With limits, we partition an interval into manageable sequences, allowing us to shrink or expand boundaries to approach problematic points.In our integral \( \int_{0}^{\pi} \frac{d x}{\cos x - 1} \), the singular points \( x = 0 \) and \( x = \pi \) are approached using limits:

• We set limits by evaluating the behavior of the integral as \( x \to 0^+ \) and \( x \to \pi^- \).
• Limits help us delineate parts of the integral that need close observation, preventing direct calculation issues due to singularities.

Practicing with limits, in this case, \( \lim_{a \to 0^+} \int_{a}^{\frac{\pi}{2}} \frac{d x}{\cos x-1} + \lim_{b \to \pi^-} \int_{\frac{\pi}{2}}^{b} \frac{d x}{\cos x-1} \), facilitates understanding how the integral behaves at boundaries. Evaluations show divergence due to infinite approaching trends as \( x \to 0^+ \) and \( x \to \pi^- \). This systematic approach secures accurate deductions on the integral's nature.

Evaluating Integrals

When evaluating improper integrals, it’s essential to prepare for more complex behavior than seen with regular integrals. The given integral \( \int_{0}^{\pi} \frac{d x}{\cos x - 1} \) cannot be straightforwardly calculated due to its improper nature. To evaluate, we examine the form of the integrand: \( \frac{1}{\cos x - 1} \) needs rewriting, often revealing points of discontinuity.

• Firstly, rewrite as \( -\frac{1}{2\sin^2(\frac{x}{2})} \) to gain clarity on where values are undefined.
• Identify key points where the function becomes problematic, here at \( x = 0 \) and \( x = \pi \).

By dividing the integral and applying limits, we can assess behavior at these points. It becomes evident that the function grows without bounds, indicating divergence.Effective strategies for improper integrals may include:

• Substitution and transformation to mitigate difficult parts.
• Using numerical integration when analytical solutions become too complex.

Evaluating an integral with possible divergence requires attentiveness, ensuring we thoroughly test endpoints and apply limits appropriately.