Improper Integral

Definitions

$f$ is defined on	$f$ is integrable on every interval	The improper integral of $f$
$(a,b]$	$[a,t]$ where $a<t<b$	${\int }_a^{b}f(x)dx=\underset{t\to b^{-}}{\lim }{\int }_a^{t}f(x)dx$
$[a,b)$	$[t,b]$ where $a<t<b$	${\int }_a^{b}f(x)dx=\underset{t\to a^{+}}{\lim }{\int }_t^{b}f(x)dx$
$[a,\infty )$	$[a,t]$ where $a<t$	${\int }_a^{\infty }f(x)dx=\underset{t\to \infty }{\lim }{\int }_a^{t}f(x)dx$
$(-\infty ,b]$	$[t,b]$ where $t<b$	${\int }_{-\infty }^{b}f(x)dx=\underset{t\to -\infty }{\lim }{\int }_t^{b}f(x)dx$

• If the limit exists (finite), then the improper integral converges; otherwise, it diverges.
• Given two improper integrals on intervals with a common endpoint (i.e. $[a,c)$ and $(c,b]$, or $(a,c]$ and $[c,b)$, or $(-\infty ,c]$ and $[c,\infty )$, or $[a,c]$ and $[c,\infty )$, etc..)
  • The sum of the two improper integrals is the improper integral of $f$ on the union of the two intervals, and it converges if both of the two improper integrals converge; otherwise, it diverges.
  • If the sum converges, then it is equal to the sum of the two integrals, and it is denoted as ${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$. (similar for other cases)
  • The convergence of the sum and its value are independent of the choice of $c$. (see 3.8, 3.13)

Theorems

• (3.3) if $f$ is integrable on $[a,b]$ then the improper integral over $(a,b]$ (or $[a,b)$ or $(a,b)$) is equal to the integral over $[a,b]$
• (q3.22)
  • $f$ is defined on $[a,\infty )$ and integrable on every finite subinterval of $[a,\infty )$, and $F^{\prime }=f$, then: ${\int }_a^{\infty }f(x)dx$ converges iff $\underset{x\to \infty }{\lim }F(x)$ exists.
    • in such case ${\int }_a^{\infty }f(x)dx=\underset{x\to \infty }{\lim }(F(x))-F(a)$
  • $f$ is defined on $\mathbb{R}$, and $F^{\prime }=f$, then: ${\int }_{-\infty }^{\infty }f(x)dx$ converges iff $\underset{x\to \infty }{\lim }F(x)$ and $\underset{x\to -\infty }{\lim }F(x)$ exist.
    • in such case ${\int }_{-\infty }^{\infty }f(x)dx=\underset{x\to \infty }{\lim }F(x)-\underset{x\to -\infty }{\lim }F(x)$
• (q3.33) If ${\int }_a^{\infty }f(x)dx$ converges and $\underset{x\to \infty }{\lim }f(x)$ exists then $\underset{x\to \infty }{\lim }f(x)=0$

Examples

• (e3.11a) ${\int }_1^{\infty }\frac{\sin x}{x^2}dx$ conv.
• (e3.11b) ${\int }_1^{\infty }\frac{p(x)}{e^x}dx$ converge for every polynomial $p$

Convergence Tests

p-test

• (3.2,q3.5) The integrals ${\int }_0^b\frac{dx}{x^p}$ and ${\int }_a^b\frac{dx}{(x-a{)}^p}$ and ${\int }_a^b\frac{dx}{(b-a{)}^p}$ converges iff $p<1$
• (3.12) if $a>0$ then ${\int }_a^{\infty }\frac{dx}{x^p}$ converges iff $p>1$
• ${\int }_1^{\infty }\frac{dx}{x^p}dx=\{\begin{array}{ll}\frac{1}{p-1}& p>1\\ \text{diverges}& p\le 1\end{array}$

Cauchy’s Criterion

• (3.4) Let $f(x)$ be defined on $(a,b]$ and integrable on every interval $[t,b]$ (where $a<t<b$). Then:
  • The improper integral ${\int }_a^{b}f(x)dx$ converges iff $\forall \epsilon >0,\exists \delta >0:\forall r,s\in [a,a+\delta ],\left|{\int }_r^{s}f(x)dx\right|<\epsilon$
• (3.15) Let $f(x)$ be defined on $[a,\infty )$ and integrable on every interval $[a,t]$ (where $a<t$). Then:
  • The improper integral ${\int }_a^{\infty }f(x)dx$ converges iff $\forall \epsilon >0,\exists M>a:\forall r,s\in [M,\infty ),\left|{\int }_r^{s}f(x)dx\right|<\epsilon$
• (similar for $[a,b)$ and $(-\infty ,b]$)

Comparison Tests $(a,b]$

Let $f,g$ be non-negative functions defined on $(a,b]$ and integrable on every closed subinterval of $(a,b]$.

Direct Comparison Test

• (3.5) If $\exists \delta >0:\forall x\in (a,a+\delta ),0\le f(x)\le g(x)$ then:
  • if ${\int }_a^{b}g$ converges, then ${\int }_a^{b}f$ converges
  • if ${\int }_a^{b}f$ diverges, then ${\int }_a^{b}g$ diverges
• (similar test for $[a,b)$)

Limit Comparison Test

• (3.5*) Given $\underset{x\to a^{+}}{\lim }\frac{f(x)}{g(x)}=L$ exists, then:
  • if $0<L<\infty$ then ${\int }_a^{b}g$ converges iff ${\int }_a^{b}f$ converges
  • if $L=0$ and ${\int }_a^{b}g$ converges, then ${\int }_a^{b}f$ converges
  • if $L=\infty$ and ${\int }_a^{b}g$ diverge, then ${\int }_a^{b}f$ diverge
  • (similar test for $[a,b)$)

Comparison Tests $[a,\infty )$

Let $f,g$ be non-negative functions defined on $[a,\infty )$ and integrable on every closed subinterval $[a,t]\subset [a,\infty )$.

Direct Comparison Test

• (3.16) If $\exists A\ge a:\forall x\in [A,\infty ),0\le f(x)\le g(x)$ then:
  • if ${\int }_a^{\infty }g$ converges, then ${\int }_a^{\infty }f$ converges
  • if ${\int }_a^{\infty }f$ diverges, then ${\int }_a^{\infty }g$ diverges

Limit Comparison Test

• (3.16-*) Given $\underset{x\to \infty }{\lim }\frac{f(x)}{g(x)}=L$ exists (finite or infinite) then:
  • if $0<L<\infty$ then ${\int }_a^{b}g(x)dx$ converges if and only if ${\int }_a^{b}f(x)dx$ converges
  • if $L=0$ and ${\int }_a^{b}g(x)dx$ converges then ${\int }_a^{b}f(x)dx$ converges
  • if $L=\infty$ and ${\int }_a^{b}g(x)dx$ diverge then ${\int }_a^{b}f(x)dx$ diverge

Dirichlet’s Test

• (3.19) Let $f$ and $g$ be continuous on $[a,\infty )$
  • If the following conditions hold:
    • $f(x)$ is decreasing on $[a,\infty )$, and $\underset{x\to \infty }{\lim }f(x)=0$
    • $f(x)$ continuously differentiable on $[a,\infty )$
    • The function $G(x)={\int }_a^{x}g(t)dt$ is bounded on $[a,\infty )$
  • Then ${\int }_a^{\infty }f(x)g(x)dx$ converges

Integral Test

Integral Test

(5.19)

• Given a series $\sum a_k$ of nonnegative terms, where $a_k$ is decreasing.
• Given $f(x)$ is decreasing (weakly), nonnegative on $[1,\infty )$, and integrable every fintie interval. And $a_n=f(n)$ for all $n\in \mathbb{N}$.

The series $\sum a_k$ converges if and only if the improper integral ${\int }_1^{\infty }f(x)dx$ converges.

We can use integral test also when $f$ is decreasing and nonnegative on $[k_0,\infty )$ for some $k_0\in \mathbb{N}$. In this case, the series ${\sum }_{k=k_0}^{\infty }{a}_k$ converges if and only if the improper integral ${\int }_{k_0}^{\infty }f(x)dx$ converges. Hence, the series ${\sum }_{k=1}^{\infty }{a}_k$ converges if and only if the series ${\sum }_{k=k_0}^{\infty }{a}_k$ converges.

Link to original

Absolutely Integrable Function

• (d3.6) Let $f(x)$ defined on $(a,b]$ and integrable in every closed subinterval, we say that $f(x)$ is absolutely integrable on $(a,b]$ if ${\int }_a^b\left|f(x)\right|dx$ is integrable in that case we say that ${\int }_a^{b}f(x)dx$ absolutely converges (similar for $[a,b)$
• (3.7) if $f$ is absolutely integrable on $(a,b]$ then $f$ is integrable on $(a,b]$. (similar for $[a,b)$)
• (d3.17) Let $f(x)$ defined on $[a,\infty )$ and integrable in every closed subinterval $[a,t]\subset [a,\infty )$, we say that $f(x)$ is absolutely integrable on $[a,\infty )$ if ${\int }_a^{\infty }\left|f(x)\right|dx$ converges.
  • In that case we say that ${\int }_a^{\infty }f(x)dx$ absolutely converges.
  • Otherwise, (i.e., ${\int }_a^{\infty }f(x)dx$ converges but ${\int }_a^{\infty }\left|f(x)\right|dx$ diverges) we say that ${\int }_a^{\infty }f(x)dx$ conditionally converges.
• (3.18) if $f$ is absolutely integrable on $[a,\infty )$ then $f$ is integrable on $[a,\infty )$