Chapter 4: Integration
Section 4.5: Improper Integrals
Show that ∫1∞1xs ⅆx = 1s−1s>1∞s≤1.
For the case s>1:
∫1∞1xs ⅆx=limt→∞∫1tx−s ⅆx = limt→∞x−s+1−s+11t = 11−s limt→∞t1−s−1 = −11−s=1s−1
Since s>1, the exponent 1−s is negative, so t1−s→0 as t→∞.
In Example 4.5.2 where s=1, the integral has been shown to diverge.
For the case s<1:
∫1∞1xs ⅆx=limt→∞∫1tx−s ⅆx = limt→∞x−s+1−s+11t = 11−s limt→∞t1−s−1= ∞
Since s<1, the exponent 1−s is positive, so t1−s→∞ as t→∞.
Unfortunately, it is no longer possible to evaluate this integral in a syntax-free way via the Context Panel. One must use the "assuming" option explicitly.
Integrate to a finite endpoint
Control-drag the integral and edit the upper limit from ∞ to t
Context Panel: Simplify≻Assuming Real Range (See Figure 4.5.1(a).)
Context Panel: Assign to a Name≻q
∫1t1xs ⅆx assuming x>1,x<t = −−1+t−s+1s−1→assign to a nameq
The limit when s>1r
The limit when s≤1
Expression palette: Limit template
Fill in the template and append the assuming command.
Context Panel: Evaluate and Display Inline
limt→∞q assuming s>1 = 1s−1
limt→∞q assuming s≤1 = ∞
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