Chapter 5: Q. 41 (page 478)

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x$

Short Answer

Expert verified

The value of the integral is $\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x = 0 .$

Step by step solution

Step 1. Given information.

The given integral is the following.

$\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x$

Step 2. Value of integral.

solve this improper integral by splitting at $x = 0 .$

$\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x = \int_{- \infty}^{0} \frac{x}{{(x^{2} + 1)}^{2}} d x + \int_{0}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x = \lim_{b \to - \infty} \int_{b}^{0} \frac{x}{{(x^{2} + 1)}^{2}} d x + \lim_{a \to \infty} \int_{0}^{a} \frac{x}{{(x^{2} + 1)}^{2}} d x = \lim_{b \to - \infty} {[- \frac{1}{2} {(x^{2} + 1)}^{- 1}]}_{b}^{0} + \lim_{a \to \infty} {[- \frac{1}{2} {(x^{2} + 1)}^{- 1}]}_{0}^{a} = \lim_{b \to - \infty} [\frac{- 1}{2} - \frac{- 1}{2} {(b^{2} + 1)}^{- 1}] + \lim_{a \to \infty} [- \frac{1}{2} {(a^{2} + 1)}^{- 1} - (\frac{- 1}{2})] = (\frac{- 1}{2} - 0) (0 + \frac{1}{2}) = 0$

So The value of the integral is $\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x = 0 .$

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of integral.

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.∫-∞∞xx2+12dx

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of integral.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

Mechanics Maths

Applied Mathematics

Decision Maths

Statistics

Study anywhere. Anytime. Across all devices.

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{- \infty}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x$