Calculus- Early Transcendentals, 2021a

7.7. Improper Integrals 297
Example 7.53: Comparison Test
∫ ∞ cos 2 x
Show that
2 x 2 dx converges.
Solution. We use the comparison test to show that it converges. Note that 0 ≤ cos 2 x ≤ 1 and hence
0 ≤ cos2 x
x 2 ≤ 1 x 2 .
∫ ∞
Thus, taking f (x)=1/x 2 and g(x)=cos 2 x/x 2 1
we have f (x) ≥ g(x) ≥ 0. One can easily see that
2 x 2 dx
∫ ∞ cos 2 x
converges. Therefore,
x 2 dx also converges. ♣
Exercises for Section 7.7
2
Exercise 7.7.1 Determine whether
Exercise 7.7.2 Determine whether
∫ ∞
1
∫ ∞
Exercise 7.7.3 Evaluate the improper integral
Exercise 7.7.4 Determine if
Exercise 7.7.5 Show that
Exercise 7.7.6 Evaluate
∫ ∞
0
∫ ∞
−∞
∫ e
e
1
dx is convergent or divergent.
x2 1
x √ dx is convergent or divergent.
lnx
∫ ∞
0
e −3x dx.
1
dx is convergent or divergent. Evaluate it if it is convergent.
1 x(lnx) 2 (
e −x sin 2 πx
)
dx converges.
2
1
x 2 + 1 dx and ∫ ∞
−∞
x
x 2 + 1 dx.
Exercise 7.7.7 Determine whether the following improper integrals are convergent or divergent. Evaluate
those that are convergent.
(a) ∫ ∞
0
1
x 2 + 1 dx
(b) ∫ ∞
0
x
x 2 + 1 dx
(c) ∫ ∞
0 e −x (cosx + sinx)dx. [Hint: What is the derivative of −e −x cosx?]
(d) ∫ π/2
0
sec 2 xdx