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Chapter 12: Problem 18

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{-1} \frac{1}{x^{2}} d x $$

Short Answer

Expert verified
The given improper integral converges and its value is 1.

Step by step solution

01

Investigate Convergence or Divergence

To determine whether the integral converges or diverges, a limit substitution is made. Here, let's let b be a real number such that -鈭� < b < -1. So, we first consider the integral \(\int_{b}^{-1} \frac{1}{x^{2}} dx\).
02

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is used to evaluate the integral, thus we have \(-\[ \frac{1} {x} \]_{b}^{-1} = -(-1 - - \frac{1} {b}) = 1 - \frac{1} {b}\).
03

Taking limit as b approaches -鈭�

Now, the limit is taken as \(b\) approaches \(-鈭瀄), which gives us \(\lim_{b \to -\infty} (1 - \frac{1} {b})\). Given that the limit of a sum is equal to the sum of the limits (and the fact that the limit of a constant is that constant), this limit simplifies to \(1 - 0 = 1\).
04

Concretizing the Result

Since the limit is finite, the improper integral is convergent and its value equals the established limit, which in this case is 1.

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