Proper and Improper Integrals
Math 122C – Frank Swenton, Spring 2017
Basic Integration
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• Indefinite integrals: f (x) dx is a function, an antiderivative of f .
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• Definite integrals: a f (x) dx is a number, the “product” (b − a) times f
(or, in geometric terms, the area under y = f (x) from x = a . . . b).
b
Z
Z b
f (x) dx ,
• These are related by the FTC:
f (x) dx =
a
a
if f is defined and continuous on the finite interval [a, b].
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• Properties of integrals:
f (x) ± g(x) dx = f (x) dx ± g(x) dx
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k · f (x) dx = k · f (x) dx
• Some basic antiderivatives (these all come from
Z
xa+1
xa dx =
+ C (if a 6= −1)
a+1
Z
sin x dx = − cos x + C
Z
sec2 x dx = tan x + C
Z
tan x dx = ln | sec x| + C
Z
ex dx = ex + C
derivative formulae):
Z
1
dx = ln |x| + C
x
Z
cos x dx = sin x + C
Z
sec x tan x dx = sec x + C
Z
sec x dx = ln | sec x + tan x| + C
• Remember to write powers of x in the form xa before using the power rule.
Improper Integrals – definitions
• A proper integral is a definite integral where the interval is finite and the integrand is defined
and continuous at all points in the interval.
– Proper integrals always converge, that is, always give a finite value
• The trouble spots of a definite integral are the points in the interval of integration that make it
an improper integral, i.e., keep it from being proper. They are of two types:
a. Points at which the integrand is undefined or discontinuous
b. ∞ and −∞ are always trouble spots when they appear as limits of integration
• A simple improper integral is an improper integral with only one trouble spot, that trouble spot
being at an endpoint of the interval. Simple improper integrals are defined to be the appropriate
Z 1
Z 1
limits of proper integrals, e.g.:
1
1
dx = lim
dx
+
ε→0
0 x
ε x
– If the limit exists as a real number, then the simple improper integral is called convergent.
– If the limit doesn’t exist as a real number, the simple improper integral is called divergent.
Dealing with improper integrals
• First, locate all trouble spots, and split the integral into simple improper integrals (a sketch of the
interval may be helpful), then deal with each piece in turn.
– If each and every one of the pieces converges, the original integral converges to their sum
– If even just one of the pieces diverges, the original integral diverges