Calculus- Early Transcendentals, 2021a

292 Techniques of Integration
Therefore, the integral is divergent.
♣
Example 7.46: Improper Integral
Determine whether
∫ ∞
−∞
Solution. We must compute both
xsin(x 2 )dx is convergent or divergent.
∫ ∞
0
xsin(x 2 )dx and
∫ 0
−∞
xsin(x 2 )dx. Note that we don’t have to split the
integral up at 0, any finite value a will work. First we compute the indefinite integral. Let u = x 2 ,then
du = 2xdx and hence, ∫
xsin(x 2 )dx = 1 ∫
sin(u)du = − 1 2
2 cos(x2 )+C
Using the definition of improper integral gives:
∫ ∞
]∣ ∣∣∣
xsin(x 2 R
)dx = lim
0
∫ R
R→∞ 0
[
xsin(x 2 )dx = lim − 1
R→∞ 2 cos(x2 )
0
= − 1 2 lim
R→∞ cos(R2 )+ 1 2
This limit does not exist since cosx oscillates between ∫ −1 and+1. In particular, cosx does not approach
∞
∫ ∞
any particular value as x gets larger and larger. Thus, xsin(x 2 )dx diverges, and hence, xsin(x 2 )dx
diverges.
0
−∞
♣
When there is a discontinuity in [a,b] or at an endpoint, then the improper integral is as follows.
Definition 7.47: Definitions for Improper Integrals
If f (x) is continuous on (a,b], then the improper integral of f over (a,b] is:
∫ b
a
∫ b
f (x)dx := lim f (x)dx.
R→a + R
If f (x) is continuous on [a,b), then the improper integral of f over [a,b) is:
∫ b
a
∫ R
f (x)dx := lim f (x)dx.
R→b − a
If the limit above exists and is a finite number, we say the improper integral converges. Otherwise,we
say the improper integral diverges.
When there is a discontinuity in the interior of [a,b], we use the following definition.
Definition 7.48: Definitions for Improper Integrals
If f has a discontinuity at x = c where c ∈ [a,b], and both both
convergent, then f over [a,b] is:
∫ b
∫ c ∫ b
f (x)dx := f (x)dx+ f (x)dx
a
a
c
∫ c
a
∫ b
f (x)dx and f (x)dx are
c