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Improper Integrals I
Improper Integrals I
Improper Integrals I by Mika Seppälä
Improper Integrals
• An integral is improper if either:
• the interval of integration is infinitely long or
• if the function has singularities in the interval of integration (or both).
DefinitionDefinition
f x( )dx
a
b
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
• Improper integrals cannot be
defined as limits of Riemann sums. Neither
can one approximate them numerically
using methods based on evaluating
Riemann sums.
f x( )dx
a
b
∫
Improper Integrals I by Mika Seppälä
The integral is improper because
the interval of integration is infinitely long.
e−x dx0
∞
∫1
IMPROPER INTEGRALS
ExamplesExamples
Improper Integrals I by Mika Seppälä
is improper because the integrand has a singularity.
dx
x20
1
∫
ExamplesExamples
2
IMPROPER INTEGRALS
Improper Integrals I by Mika Seppälä
is improper because the integrand has a singularity and the interval of integration is infinitely long.
ExamplesExamples
3
e−x
xdx
0
∞
∫
IMPROPER INTEGRALS
Improper Integrals I by Mika Seppälä
Improper Integrals
DefinitionDefinition
Assume that the function f takes finite values on the interval [a, ∞).
If the limit exists and is finite, the improper integral converges, and
limb→ ∞
f x( )dxa
b
∫
f x( )dx
a
∞
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
ExampleExample
e−x dx0
∞
∫ =limb→ ∞
e−x dx0
b
∫
=lim
b→ ∞−e−x⎡
⎣⎤⎦0
b
=lim
b→ ∞−e−b − −e0
( )( ) =1.
Hence the integral converges.
e−x dx0
∞
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
DefinitionDefinition
Assume that the function f takes finite values on the interval [a, ∞).
If the limit does not exists or is not finite, the improper integral diverges
limb→ ∞
f x( )dxa
b
∫
f x( )dx
a
∞
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
ExampleExample
ex dx0
∞
∫ =limb→ ∞
ex dx0
b
∫
=lim
b→ ∞ex⎡
⎣⎤⎦0
b
=lim
b→ ∞eb −e0
( ) =∞.
Hence the integral diverges.
ex dx0
∞
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
DefinitionDefinition
Assume that the function f has a singularity at x = a. If the limit exists and is finite, the improper integral converges, and
limα→ a+
f x( )dxα
b
∫
f x( )dx
a
b
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
ExampleExample
dx
x0
1
∫ = limα→ 0 +
dx
xα
1
∫
= limα→ 0 +
x
2
⎡
⎣⎢⎢
⎤
⎦⎥⎥α
1
= limα→ 0 +
12
−α2
⎛
⎝⎜
⎞
⎠⎟
=
12
.
Hence the integral converges.
dx
x0
1
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
DefinitionDefinition
Assume that the function f has a singularity at x = a.
If the limit does not exist or is not finite, the improper integral diverges.
limα→ a+
f x( )dxα
b
∫
f x( )dx
a
b
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
ExampleExample
dx
x0
1
∫ = limα→ 0 +
dx
xα
1
∫
= lim
α→ 0 +ln x⎡
⎣⎤⎦α
1
= lim
α→ 0 +ln1 −lnα( ) =∞.
Hence the integral diverges.
dx
x0
1
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
DefinitionDefinition
If the function f has a singularity at a point c,
a < c < b, then the improper integral
converges if and only if both improper
integrals and converge.
f x( )dx
a
b
∫
f x( )dx
a
c
∫
f x( )dxc
b
∫
In this case
f x( )dxa
b
∫ = f x( )dxa
c
∫ + f x( )dxc
b
∫ .
Improper Integrals I by Mika Seppälä
Improper Integrals
ExampleExample
dx
x3−1
1
∫ =dx
x3−1
0
∫ +dx
x30
1
∫
= lim
β→ 0 −
dx
x3−1
β
∫ + limα→ 0 +
dx
x3α
1
∫
= limβ→ 0 −
32
x23
⎡
⎣⎢⎢
⎤
⎦⎥⎥−1
β
+ limα→ 0 +
32
x23
⎡
⎣⎢⎢
⎤
⎦⎥⎥α
1
= lim
β→ 0 −
32
β23 −
32
⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜
⎞
⎠⎟ + lim
α→ 0 +
32
−32
α23
⎛
⎝⎜
⎞
⎠⎟
=−
32
+32
=0 .
Hence the integral converges.
dx
x3−1
1
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
DefinitionDefinition
If the function f has a singularity at a point c,
a < c < b, then the improper integral
diverges if either or diverges.
f x( )dx
a
b
∫
f x( )dx
a
c
∫
f x( )dxc
b
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
ExampleExample
= lim
β→ 0 −
dx
x3−1
β
∫ + limα→ 0 +
dx
x3α
1
∫ = lim
β→ 0 −
1−2
x−2⎡
⎣⎢
⎤
⎦⎥−1
β
+ limα→ 0 +
1−2
x−2⎡
⎣⎢
⎤
⎦⎥α
1
= lim
β→ 0 −−
12
β−2 − −12
⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜
⎞
⎠⎟ + lim
α→ 0 +−
12
− −12
α−2⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜
⎞
⎠⎟.
Neither limits exists. The integral diverges.
dx
x3−1
1
∫
Improper Integrals I by Mika Seppälä
Improper Integrals
WarningWarning The integral diverges.
dx
x3−1
1
∫
Trying to compute that integral by the Fundamental Theorem of Calculus, one gets
dx
x3−1
1
∫ =1−2
x−2
−1
1
=−
12
− −12
⎛
⎝⎜⎞
⎠⎟=0.
This is an incorrect computation.
Improper Integrals I by Mika Seppälä
Summary
• An integral is improper if either: the interval of integration is infinitely long or if the function has singularities in the interval of integration (or both). Such integrals cannot be defined as limits of Riemann sums. They must be defined as limits of integrals over finite intervals where the function takes only finite values.
