Chapter 4: Integration
Section 4.5: Improper Integrals
Example 4.5.3
Evaluate the improper integral ∫0∞11+x2 ⅆx.
Solution
Mathematical Solution
∫0∞11+x2 ⅆx
=limt→∞∫0t11+x2 ⅆx
=limt→∞arctanx0t
=limt→∞arctant−arctan0
=limt→∞arctant−0
=limt→∞arctant
=π/2
Maple Solution
Apply Maple to the improper integral
Control-drag the integral. Context Panel: Evaluate and Display Inline
∫0∞11+x2 ⅆx = 12π
Integrate to a finite endpoint, then take the limit
Control-drag the integral Change the upper limit from ∞ to t
Context Panel: Evaluate and Display Inline
∫0t11+x2 ⅆx = arctant
Expression palette: Limit template Context Panel: Evaluate and Display Inline
limt→∞arctant = 12π
The astute reader will note the difference in Maple's behavior here as opposed to that in Examples 4.5.1 and 4.5.2. In those examples, the integrands have a vertical asymptote at x=0, whereas this present integrand does not. In the first two examples, if t<0, the integration is from x=1 back across this vertical asymptote, an integration that is not permitted. Hence, in those two examples, Maple must know somehow that t>0.
<< Previous Example Section 4.5 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
