Explanation Given: The integral: I = ∫ − ∞ ∞ d x ( x 2 + 1 ) 3 2 Formula used: I = ∫ a b f ( x ) d x = [ F ( x ) ] a b = F ( b ) − F ( a ) I = ∫ a b f ( x ) d x = ∫ a c 1 f ( x ) d x + ∫ c 1 c 2 f ( x ) d x + ... + ∫ c n − 1 b f ( x ) d x , Where a < c 1 < c 2 < ... < c n − 1 < b ln 1 = 0 Calculation: The integral we have I = ∫ − ∞ ∞ d x ( x 2 + 1 ) 3 2 Let x = tan θ ⇒ d x = sec 2 θ d θ Also, x = ∞ ⇒ tan θ = ∞ ⇒ θ = π 2 x = − ∞ ⇒ tan θ = − ∞ ⇒ θ = − π 2 ∴ I = ∫ − ∞ ∞ d x ( x 2 + 1 ) 3 2 = ∫ − π 2 π 2 sec 2 θ d θ ( tan 2 θ + 1 ) 3 2 = ∫ − π 2 π 2 sec 2

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 8.7, Problem 48E

To determine

The value of the integral

$I = \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} + 1)}^{\frac{3}{2}}}$

Whether the given integral is convergent.

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Chapter 8.7, Problem 47E

Chapter 8.7, Problem 49E

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Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The value of the integral I = ∫ − ∞ ∞ d x ( x 2 + 1 ) 3 2 Whether the given integral is convergent.

Solution Summary: The author explains the formula used to determine the value of the integral, which is convergent.

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The value of the integral

$I = \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} + 1)}^{\frac{3}{2}}}$

Whether the given integral is convergent.

Want to see the full answer?

Chapter 8 Solutions

The value of the integral I = ∫ − ∞ ∞ d x ( x 2 + 1 ) 3 2 Whether the given integral is convergent.

Solution Summary: The author explains the formula used to determine the value of the integral, which is convergent.

Concept explainers

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The value of the integral I=∫−∞∞dx(x2+1)32 Whether the given integral is convergent.

Want to see the full answer?

Chapter 8 Solutions

The value of the integral

$I = \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} + 1)}^{\frac{3}{2}}}$

Whether the given integral is convergent.