MATHS 244 ANALYSIS
Improper Integrals
Improper integrals extend the idea of definite integrals to cases where the interval of
integration is infinite, or the integrand becomes unbounded within the interval.
We can think of these as limits of ordinary integrals.
Definition of Improper Integrals (Type I)
These occur when the interval of integration is infinite.
Case 1: Infinite upper limit
If 𝑓(𝑥) is continuous on [𝑎, ∞), then we define:
∞ 𝑡
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑡→∞ 𝑎
• If the limit exists (finite number), the integral converges.
• If the limit does not exist (approaches infinity or oscillates), it diverges.
Case 2: Infinite lower limit
If 𝑓(𝑥) is continuous on (−∞, 𝑏], then:
𝑏 𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥
−∞ 𝑡→−∞ 𝑡
Case 3: Both limits infinite
If both integrals
∞ 𝑎
∫𝑎 𝑓(𝑥) 𝑑𝑥and ∫−∞ 𝑓(𝑥) 𝑑𝑥 converge for some finite 𝑎, then:
∞ 𝑎 ∞
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
−∞ −∞ 𝑎
Example:
∞ 𝑡
1 1 1
∫ 𝑑𝑥 = lim ∫ 𝑑𝑥 = lim (1 − 𝑡 ) = 1 Converges
𝑥2 𝑡→∞ 1 𝑥2 𝑡→∞
1
∞
1
∫ 𝑑𝑥 = lim ln 𝑡 = ∞ Diverges
𝑥 𝑡→∞
1
,Definition of Improper Integrals (Type II)
These occur when 𝑓(𝑥)becomes unbounded at a point within a finite interval.
Case 1: Unbounded near the upper limit 𝒃
If 𝑓is continuous on [𝑎, 𝑏) but has a vertical asymptote at 𝑏:
𝑏 𝑡
∫ 𝑓(𝑥) 𝑑𝑥 = lim− ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑡→𝑏 𝑎
Case 2: Unbounded near the lower limit 𝒂
If 𝑓 is continuous on (𝑎, 𝑏] but unbounded near 𝑎:
𝑏 𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = lim+ ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑡→𝑎 𝑡
Example:
1 1
1
∫ 𝑑𝑥 = lim+ ∫ 𝑥 −1/2 𝑑𝑥 = lim+ [2(1 − √𝑡)] = 2
0 √𝑥 𝑡→0 𝑡 𝑡→0
Convergent.
1
1
∫ 𝑑𝑥 = lim+ ln(1/𝑡) = ∞
0 𝑥 𝑡→0
Divergent.
Type III Improper Integrals
These combine Type I and Type II situations — when the function is unbounded and the
interval is infinite.
If 𝑓 is unbounded near 𝑎 and the domain extends to infinity:
∞ 𝑐 ∞
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑎 𝑐
𝑐 ∞
where both the Type II integral ∫𝑎 𝑓(𝑥) 𝑑𝑥 and Type I integral ∫𝑐 𝑓(𝑥) 𝑑𝑥 must converge.
,Convergence Conditions and Lemmas
Lemma 1 (Shift Independence)
∞
If 𝑓(𝑥) is continuous on [𝑎, ∞)and ∫𝑎 𝑓(𝑥) 𝑑𝑥converges, then for any 𝑏 > 𝑎,
∞
∫𝑏 𝑓(𝑥) 𝑑𝑥 also converges.
So convergence depends only on the behaviour at infinity, not where you start.
Lemma 2 (Linearity)
∞ ∞
If both ∫𝑎 𝑓(𝑥) 𝑑𝑥 and ∫𝑎 𝑔(𝑥) 𝑑𝑥 converge, then for any constants 𝛼, 𝛽 ∈ ℝ:
∞
∫ [𝛼𝑓(𝑥) + 𝛽𝑔(𝑥)]𝑑𝑥
𝑎
also converges.
Lemma 3 (p–Test for Convergence)
∞
1
∫ 𝑑𝑥
𝑏 (𝑥 − 𝑎)𝑝
• Convergent if 𝑝 > 1
• Divergent if 𝑝 ≤ 1
This gives a simple test for rational functions with polynomial denominators.
Lemma 4 (Exponential Test)
∞
∫ 𝑒 −𝑝𝑥 𝑑𝑥
𝑎
• Convergent if 𝑝 > 0
• Divergent if 𝑝 ≤ 0
Because exponential decay dominates polynomial growth.
Lemma 5 (Singular Endpoint Test)
𝑏 𝑏
1 1
∫ 𝑝
𝑑𝑥 or ∫ 𝑝
𝑑𝑥
𝑎 (𝑥 − 𝑎) 𝑎 (𝑏 − 𝑥)
• Convergent if 𝑝 < 1
• Divergent if 𝑝 ≥ 1
, Convergence Rules
Form of f(x) Converges when Diverges when
1
as 𝑥 → ∞ 𝑝>1 𝑝≤1
𝑥𝑝
1
as 𝑥 → 0+ 𝑝<1 𝑝≥1
𝑥𝑝
𝑒 −𝑝𝑥 as 𝑥 → ∞ 𝑝>0 𝑝≤0
Comparison & Quotient Tests
When you can’t directly integrate a function to test convergence, you compare it to
another function whose behaviour you already know.
These tests are essential for determining whether an improper integral converges or
diverges without evaluating it explicitly.
The Comparison Test (Type I Integrals)
Applies when the interval is infinite — i.e. the issue is the upper or lower limit extending
to ∞ or −∞.
Setup
Let 𝑓(𝑥) and 𝑔(𝑥) be continuous functions on [𝑎, ∞) such that
0 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥)
for all 𝑥 ≥ 𝑎.
Rules
∞
1. If ∫𝑎 𝑓(𝑥) 𝑑𝑥 converges,
∞
then ∫𝑎 𝑔(𝑥) 𝑑𝑥also converges.
∞
2. If ∫𝑎 𝑔(𝑥) 𝑑𝑥 diverges,
∞
then ∫𝑎 𝑓(𝑥) 𝑑𝑥 also diverges.
Intuition:
A smaller function can’t “blow up” if a bigger one stays finite.
Likewise, if a smaller one already diverges, a larger one must also.
Example 1 Convergent Comparison
1 1
𝑓(𝑥) = 2
, 𝑔(𝑥) = 3 , 𝑥 ≥ 1
𝑥 𝑥
∞ ∞
Since 0 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥) and ∫1 1/𝑥 2 𝑑𝑥 = 1 (convergent), then ∫1 1/𝑥 3 𝑑𝑥 also
converges.
Improper Integrals
Improper integrals extend the idea of definite integrals to cases where the interval of
integration is infinite, or the integrand becomes unbounded within the interval.
We can think of these as limits of ordinary integrals.
Definition of Improper Integrals (Type I)
These occur when the interval of integration is infinite.
Case 1: Infinite upper limit
If 𝑓(𝑥) is continuous on [𝑎, ∞), then we define:
∞ 𝑡
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑡→∞ 𝑎
• If the limit exists (finite number), the integral converges.
• If the limit does not exist (approaches infinity or oscillates), it diverges.
Case 2: Infinite lower limit
If 𝑓(𝑥) is continuous on (−∞, 𝑏], then:
𝑏 𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥
−∞ 𝑡→−∞ 𝑡
Case 3: Both limits infinite
If both integrals
∞ 𝑎
∫𝑎 𝑓(𝑥) 𝑑𝑥and ∫−∞ 𝑓(𝑥) 𝑑𝑥 converge for some finite 𝑎, then:
∞ 𝑎 ∞
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
−∞ −∞ 𝑎
Example:
∞ 𝑡
1 1 1
∫ 𝑑𝑥 = lim ∫ 𝑑𝑥 = lim (1 − 𝑡 ) = 1 Converges
𝑥2 𝑡→∞ 1 𝑥2 𝑡→∞
1
∞
1
∫ 𝑑𝑥 = lim ln 𝑡 = ∞ Diverges
𝑥 𝑡→∞
1
,Definition of Improper Integrals (Type II)
These occur when 𝑓(𝑥)becomes unbounded at a point within a finite interval.
Case 1: Unbounded near the upper limit 𝒃
If 𝑓is continuous on [𝑎, 𝑏) but has a vertical asymptote at 𝑏:
𝑏 𝑡
∫ 𝑓(𝑥) 𝑑𝑥 = lim− ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑡→𝑏 𝑎
Case 2: Unbounded near the lower limit 𝒂
If 𝑓 is continuous on (𝑎, 𝑏] but unbounded near 𝑎:
𝑏 𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = lim+ ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑡→𝑎 𝑡
Example:
1 1
1
∫ 𝑑𝑥 = lim+ ∫ 𝑥 −1/2 𝑑𝑥 = lim+ [2(1 − √𝑡)] = 2
0 √𝑥 𝑡→0 𝑡 𝑡→0
Convergent.
1
1
∫ 𝑑𝑥 = lim+ ln(1/𝑡) = ∞
0 𝑥 𝑡→0
Divergent.
Type III Improper Integrals
These combine Type I and Type II situations — when the function is unbounded and the
interval is infinite.
If 𝑓 is unbounded near 𝑎 and the domain extends to infinity:
∞ 𝑐 ∞
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎 𝑎 𝑐
𝑐 ∞
where both the Type II integral ∫𝑎 𝑓(𝑥) 𝑑𝑥 and Type I integral ∫𝑐 𝑓(𝑥) 𝑑𝑥 must converge.
,Convergence Conditions and Lemmas
Lemma 1 (Shift Independence)
∞
If 𝑓(𝑥) is continuous on [𝑎, ∞)and ∫𝑎 𝑓(𝑥) 𝑑𝑥converges, then for any 𝑏 > 𝑎,
∞
∫𝑏 𝑓(𝑥) 𝑑𝑥 also converges.
So convergence depends only on the behaviour at infinity, not where you start.
Lemma 2 (Linearity)
∞ ∞
If both ∫𝑎 𝑓(𝑥) 𝑑𝑥 and ∫𝑎 𝑔(𝑥) 𝑑𝑥 converge, then for any constants 𝛼, 𝛽 ∈ ℝ:
∞
∫ [𝛼𝑓(𝑥) + 𝛽𝑔(𝑥)]𝑑𝑥
𝑎
also converges.
Lemma 3 (p–Test for Convergence)
∞
1
∫ 𝑑𝑥
𝑏 (𝑥 − 𝑎)𝑝
• Convergent if 𝑝 > 1
• Divergent if 𝑝 ≤ 1
This gives a simple test for rational functions with polynomial denominators.
Lemma 4 (Exponential Test)
∞
∫ 𝑒 −𝑝𝑥 𝑑𝑥
𝑎
• Convergent if 𝑝 > 0
• Divergent if 𝑝 ≤ 0
Because exponential decay dominates polynomial growth.
Lemma 5 (Singular Endpoint Test)
𝑏 𝑏
1 1
∫ 𝑝
𝑑𝑥 or ∫ 𝑝
𝑑𝑥
𝑎 (𝑥 − 𝑎) 𝑎 (𝑏 − 𝑥)
• Convergent if 𝑝 < 1
• Divergent if 𝑝 ≥ 1
, Convergence Rules
Form of f(x) Converges when Diverges when
1
as 𝑥 → ∞ 𝑝>1 𝑝≤1
𝑥𝑝
1
as 𝑥 → 0+ 𝑝<1 𝑝≥1
𝑥𝑝
𝑒 −𝑝𝑥 as 𝑥 → ∞ 𝑝>0 𝑝≤0
Comparison & Quotient Tests
When you can’t directly integrate a function to test convergence, you compare it to
another function whose behaviour you already know.
These tests are essential for determining whether an improper integral converges or
diverges without evaluating it explicitly.
The Comparison Test (Type I Integrals)
Applies when the interval is infinite — i.e. the issue is the upper or lower limit extending
to ∞ or −∞.
Setup
Let 𝑓(𝑥) and 𝑔(𝑥) be continuous functions on [𝑎, ∞) such that
0 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥)
for all 𝑥 ≥ 𝑎.
Rules
∞
1. If ∫𝑎 𝑓(𝑥) 𝑑𝑥 converges,
∞
then ∫𝑎 𝑔(𝑥) 𝑑𝑥also converges.
∞
2. If ∫𝑎 𝑔(𝑥) 𝑑𝑥 diverges,
∞
then ∫𝑎 𝑓(𝑥) 𝑑𝑥 also diverges.
Intuition:
A smaller function can’t “blow up” if a bigger one stays finite.
Likewise, if a smaller one already diverges, a larger one must also.
Example 1 Convergent Comparison
1 1
𝑓(𝑥) = 2
, 𝑔(𝑥) = 3 , 𝑥 ≥ 1
𝑥 𝑥
∞ ∞
Since 0 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥) and ∫1 1/𝑥 2 𝑑𝑥 = 1 (convergent), then ∫1 1/𝑥 3 𝑑𝑥 also
converges.
