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Improper integrals
Improper integrals of type I and II, p-integrals, the comparison theorems.
Lectures 9–10a · Tutorial 10 (second part) · Techniques of integration and improper integrals
Textbook sections: Thomas ch. 8 Techniques of Integration (8.1–8.5, 8.8) · Stewart ch. 7 (7.1–7.5, 7.8)
Textbook sections: Thomas ch. 8 Techniques of Integration (8.1–8.5, 8.8) · Stewart ch. 7 (7.1–7.5, 7.8)
Improper integrals
Calculus I · Module 11
After this module you can
- Evaluate improper integrals as limits and decide convergence.
- Use the p-integral and comparison tests.
- Avoid the classic traps: a singularity inside the interval, symmetric limits that hide divergence, and $p=1$.
Two types
| Type | Definition |
|---|---|
| I: infinite interval | $\int_a^\infty f=\lim_{t\to\infty}\int_a^tf$ |
| II: unbounded integrand at $b$ | $\int_a^bf=\lim_{t\to b^-}\int_a^tf$ |
| singularity inside $[a,b]$ | split at $c$; BOTH parts must converge |
| whole line | $\int_{-\infty}^\infty f=\int_{-\infty}^0+\int_0^\infty$ |
Convergent = the limit is a finite number.
p-integrals and standard values
- $\int_1^\infty\frac{dx}{x^p}$ converges iff $p>1$; $\int_0^1\frac{dx}{x^p}$ converges iff $p<1$.
- $\int_2^\infty\frac{dx}{x(\ln x)^p}$ converges iff $p>1$ (value $\frac{(\ln2)^{1-p}}{p-1}$).
- $\int_{-\infty}^\infty\frac{dx}{1+x^2}=\pi$; $\int_0^1\frac{dx}{\sqrt{1-x^2}}=\frac\pi2$; $\int_0^\infty2e^{-\theta}\sin\theta\,d\theta=1$.
- $\int_0^2\frac{dx}{\sqrt{|x-1|}}=2+2=4$ (singular at $1$, split).
Comparison
- Direct: $0\le f\le g$: $\int g$ converges $\Rightarrow$ $\int f$ converges; $\int f$ diverges $\Rightarrow$ $\int g$ diverges.
- Limit comparison: $\lim\frac fg=c\in(0,\infty)$: same behaviour.
- $\int_0^\infty\frac{d\theta}{1+e^\theta}$: $\frac1{1+e^\theta}<e^{-\theta}$, converges (value $\ln2$).
- $\int_{e^e}^\infty\ln(\ln x)\,dx$: integrand $\ge1$, diverges.
Typical mistakes
Symmetric limits$\lim_{x\to\infty}\int_{-x}^x\sin t\,dt=0$ does NOT make $\int_{-\infty}^\infty\sin t\,dt$ converge.
Hidden singularityCheck for points inside the interval where the integrand blows up.
$p=1$$\int_1^\infty\frac{dx}{x}$ diverges.
Practice next
- Chapter quizzes Ch. 5 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Riemann sums, trapezoid and Simpson
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Improper integrals (Tutorial 10, Q7(a))
$\displaystyle\int_0^{\infty}2e^{-\theta}\sin\theta\,d\theta=$
Show the answer and the solution
Answer: $1$
$\int e^{-\theta}\sin\theta\,d\theta=-\frac12e^{-\theta}(\sin\theta+\cos\theta)$; from $0$ to $\infty$: $0-(-\frac12)=\frac12$, times $2$ gives $1$.
Worked example 2 · Improper integrals (Tutorial 10, Q7(b))
$\displaystyle\int_0^2\dfrac{dx}{\sqrt{|x-1|}}=$
Show the answer and the solution
Answer: $4$
Singular at $x=1$: split, $\int_0^1\frac{dx}{\sqrt{1-x}}+\int_1^2\frac{dx}{\sqrt{x-1}}=2+2=4$ (each converges since $p=\frac12<1$).
Worked example 3 · Improper integrals (Tutorial 11, Q6(a))
$\displaystyle\int_2^{\infty}\dfrac{dx}{x(\ln x)^p}$ ($p>0$) converges if and only if
Show the answer and the solution
Answer: $p>1$
$u=\ln x$: $\int_{\ln2}^\infty u^{-p}du$, a $p$-integral at infinity, finite iff $p>1$.
Worked example 4 · Improper integrals (Computation problems 10, Q7)
$\displaystyle\int_0^{\ln2}x^{-2}e^{-1/x}\,dx=$
Show the answer and the solution
Answer: $e^{- \frac{1}{\ln{\left(2 \right)}}}$
$u=-\frac1x$, $du=\frac{dx}{x^2}$: $[e^{-1/x}]_0^{\ln2}=e^{-1/\ln2}-\lim_{x\to0^+}e^{-1/x}=e^{-1/\ln2}$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Riemann sums, trapezoid and Simpson