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Chapter 6: Problem 23

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{\infty} \frac{2}{4+x^{2}} d x $$

Short Answer

Expert verified
Therefore, the improper integral converges and its value is \(\pi\).

Step by step solution

01

Breaking down the integral

Since the improper integral is from \(-\infty\) to \(\infty\), rewrite the integral as limits of two integrals: \[\lim_{a \rightarrow -\infty}\int_{a}^{0}\frac{2}{4+x^{2}} dx + \lim_{b \rightarrow \infty}\int_{0}^{b}\frac{2}{4+x^{2}} dx \]
02

Evaluating the integrals

Perform integration on both parts. The antiderivative of \(\frac{2}{4+x^{2}}\) is \( \tan^{-1}\left(\frac{x}{2}\right)\), so: \[\lim_{a \rightarrow -\infty}\left [ \tan^{-1}\left(\frac{x}{2}\right) \right ]_{a}^{0} + \lim_{b \rightarrow \infty}\left [ \tan^{-1}\left(\frac{x}{2}\right) \right ]_{0}^{b}\]
03

Apply the limits

Substitute the limits and calculate the answer: \[\lim_{a \rightarrow -\infty} \tan^{-1}\left(\frac{0}{2}\right)-\tan^{-1}\left(\frac{a}{2}\right)+ \lim_{b \rightarrow \infty} \tan^{-1}\left(\frac{b}{2}\right)-\tan^{-1}\left(\frac{0}{2}\right)\]
04

Simplifying the Equation

We find that as \(a \rightarrow -\infty\), \(\tan^{-1}\left(\frac{a}{2}\right) \rightarrow -\frac{\pi}{2}\). And as \(b \rightarrow \infty\), \(\tan^{-1}\left(\frac{b}{2}\right) \rightarrow \frac{\pi}{2}\). The answer simplifies to: \[\frac{\pi}{2} --\frac{\pi}{2}\]

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Most popular questions from this chapter

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