Problem 31 - Calculus for Biology and Medicine

Question

Determine whether $$\int_{-\infty}^{\infty} \frac{1}{x^{2}-1} d x$$ is convergent. Hint: Use the partial-fraction decomposition $$ \frac{1}{x^{2}-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) $$

Step-by-Step Solution

Answer

The integral is divergent.

Step 1: Analyze the Integrand

The function \( f(x) = \frac{1}{x^2-1} \) is defined everywhere except where the denominator is zero. Setting \( x^2 - 1 = 0 \), we find the points of discontinuity: \( x = 1 \) and \( x = -1 \).

Step 2: Apply Partial-Fraction Decomposition

Using the given partial-fraction decomposition, we rewrite the integrand as: \( \frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) \). This separation helps us understand the behavior of the integrand near its points of discontinuity.

Step 3: Setup Limit for Improper Integral Around Discontinuities

The points of discontinuity (\( x = 1 \) and \( x = -1 \)) make the integral improper. We'll evaluate the integral using limits: \[\int_{-rac{R}{2}}^{-a}\frac{1}{x^2-1}dx + \int_{-a}^{a} \frac{1}{x^2-1} dx + \int_a^{\frac{R}{2}}\frac{1}{x^2-1} dx\] and take the limit \( a \to 1^- \) and \( R \to \infty \).

Step 4: Assess Convergence of Each Integral

Each part of the integral involves expressions like \( \int \frac{1}{x-1} dx \), which leads to the natural logarithm function. For example, \( \int_{-a}^{a} \left(\frac{1}{x-1} - \frac{1}{x+1}\right) dx \) creates issues near \( x = 1 \) and \( x = -1 \), resulting in \( \ln|x-1| \) and \( \ln|x+1| \), both of which diverge as \( x \) approaches the points of discontinuity.

Step 5: Conclusion on Convergence

The divergence of the integral near \( x = 1 \) and \( x = -1 \) indicates that the overall integral:\[\int_{-rac{R}{2}}^{-a} \frac{1}{x^2-1} dx + \int_{-a}^{a} \frac{1}{x^2-1} dx + \int_a^{\frac{R}{2}} \frac{1}{x^2-1} dx \] is divergent. Since this occurs at the discontinuity points, the original integral from \(-\infty\) to \(+\infty\) is also divergent.

Key Concepts

Improper Integrals

Improper integrals are a special type of definite integrals. They occur when either the interval of integration is infinite or the function being integrated becomes infinite at some points along the interval. In the exercise given, the integral of the function \( \frac{1}{x^2-1} \) from \(-\infty\) to \(+\infty\) is improper because the function approaches infinity at \( x = -1 \) and \( x = 1 \). To handle this, we split the integral at points of discontinuity. We evaluate each segment and find the limit of the integral as we approach these points. Unfortunately, if even one segment of the integral diverges to infinity, the entire integral does too. This exercise has points where the integral diverges at both \( x = -1 \) and \( x = 1 \), leading to the conclusion that the overall integral is divergent.

Partial-Fraction Decomposition

Partial-fraction decomposition is a technique used to rewrite a complicated fraction into simpler fractions, which are easier to integrate or analyze. Here, the function \( \frac{1}{x^2-1} \) is decomposed using the identity:

• \( \frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) \)

This transformation breaks down the function into simpler parts with clear discontinuities at \( x = 1 \) and \( x = -1 \). By separating the integrand in this way, it becomes much easier to evaluate the behavior of the integral, especially around the points where the function is undefined. Partial-fraction decomposition is a valuable tool, particularly because it highlights symmetry or exposes simpler patterns that are not immediately obvious.

Convergence and Divergence

Understanding convergence and divergence is crucial when dealing with improper integrals. If the limits evaluated in an improper integral approach finite values, then the integral **converges**, meaning it has a finite solution. However, if these limits tend toward infinity, anywhere along the interval, especially near the discontinuities, the integral **diverges**, signifying there is no finite area under the curve.In the given exercise, after splitting the function into partial fractions, each separate integral was evaluated for convergence. When you try to calculate integrals such as \( \int \frac{1}{x-1} dx \) or \( \int \frac{1}{x+1} dx \), you find they relate to logarithmic functions like \( \ln|x-1| \) and \( \ln|x+1| \). As these functions approach \( x = 1 \) or \( x = -1 \), they infinitely increase, highlighting divergence.Remember, a single divergent point can determine the overall divergence of an integral. Thus, dealing with such cases requires careful assessment to confirm whether the integral converges or diverges.