7The Riemann Integral
IA Analysis I
7.2 Improper integrals
It is sometimes sensible to talk about integrals of unbounded functions or
integrating to infinity. But we have to be careful and write things down nicely.
Definition (Improper integral). Suppose that we have a function
f
: [
a, b
]
→ R
such that, for every
ε >
0,
f
is integrable on [
a
+
ε, b
] and
lim
ε→0
R
b
a+ε
f
(
x
) d
x
exists. Then we define the improper integral
Z
b
a
f(x) dx to be lim
ε→0
Z
b
a+ε
f(x) dx.
even if the Riemann integral does not exist.
We can do similarly for [a, b − ε], or integral to infinity:
Z
∞
a
f(x) dx = lim
b→∞
Z
b
a
f(x) dx.
when it exists.
Example.
Z
1
ε
x
−1/2
dx =
h
2x
−1/2
i
1
ε
= 2 −2ε
1/2
→ 2.
So
Z
1
0
x
−1/2
dx = 2,
even though x
−1/2
is unbounded on [0, 1].
Note that officially we are required to make
f
(
x
) =
x
−1/2
a function with
domain [0
,
1]. So we can assign
f
(0) =
π
, or any number, since it doesn’t matter.
Example.
Z
x
1
1
t
2
dt =
−
1
t
x
1
= 1 −
1
x
→ 1 as x → ∞
by the fundamental theorem of calculus. So
Z
∞
1
1
x
2
dx = 1.
Finally, we can prove the integral test, whose proof we omitted when we first
began.
Theorem (Integral test). Let
f
: [1
, ∞
]
→ R
be a decreasing non-negative
function. Then
P
∞
n=1
f(n) converges iff
R
∞
1
f(x) dx < ∞.
Proof. We have
Z
n+1
n
f(x) dx ≤ f(n) ≤
Z
n
n−1
f(x) dx,
since
f
is decreasing (the right hand inequality is valid only for
n ≥
2). It follows
that
Z
N+1
1
f(x) dx ≤
N
X
n=1
f(n) ≤
Z
N
1
f(x) dx + f(1)
So if the integral exists, then
P
f
(
n
) is increasing and bounded above by
R
∞
1
f(x) dx, so converges.
If the integral does not exist, then
R
N
1
f
(
x
) d
x
is unbounded. Then
P
N
n=1
f
(
n
)
is unbounded, hence does not converge.
Example. Since
R
x
1
1
t
2
dt < ∞, it follows that
P
∞
n=1
1
n
2
converges.