Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 July 2014

Lecture 15

Improper integrals

Idea: $${\int }_a^{b}f\left(x\right)dx=(\text{area under}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}x=a\hspace{0.17em}\text{to}\hspace{0.17em}x=b)\text{.}$$ $$\begin{array}{c} y=f(x) x y a b This area is ∫abf(x)dx \end{array}$$

Improper integrals: Definitions

$$\begin{array}{cc}{\displaystyle {\int }_a^{\infty }f\left(x\right)dx=\underset{b\to \infty }{\text{lim}}{\int }_a^{b}f\left(x\right)dx}& \begin{array}{c} y=f(x) a x y \end{array}\end{array}$$ and $$\begin{array}{cc}{\displaystyle {\int }_a^{b}f\left(x\right)dx=\underset{\ell \to a}{\text{lim}}{\int }_{\ell }^{b}f\left(x\right)dx\text{if}}& \begin{array}{c} y=f(x) a b x y \end{array}\end{array}$$ or $$\begin{array}{cc}{\displaystyle {\int }_a^{b}f\left(x\right)dx=\underset{\ell \to b}{\text{lim}}{\int }_a^{\ell }f\left(x\right)dx\text{if}}& \begin{array}{c} a x y \end{array}\end{array}$$

Evaluate ${\int }_0^{\infty }\frac{dx}{1+x^2}$ $$\begin{array}{c} y=11+x2 x y \end{array}$$

$$\begin{array}{ccc}{\displaystyle {\int }_0^{\infty }\frac{1}{1+x^2}dx}& =& {\displaystyle \underset{t\to \infty }{\text{lim}}{\int }_0^t\frac{1}{1+x^2}dx}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}\left({\text{tan}}^{-1}x{|}_{x=0}^{x=t}\right)}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}({\text{tan}}^{-1}t-{\text{tan}}^{-1}0)}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}({\text{tan}}^{-1}t-0)=\frac{\pi }{2}-0=\frac{\pi }{2}\text{.}}\end{array}$$

Let $p\in ℝ,$ $p>1\text{.}$ Evaluate ${\int }_1^{\infty }\frac{1}{x^p}dx\text{.}$ (Think $p=2$ for comfort.) $$\begin{array}{c} y=1x2 -1 1 x 1 y \end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_1^{\infty }\frac{1}{x^p}dx}& =& {\displaystyle \underset{t\to \infty }{\text{lim}}{\int }_1^t\frac{1}{x^p}dx}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}{\int }_1^t{x}^{-p}dx}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}\left(\frac{x^{-p+1}}{-p+1}{|}_{x=1}^{x=t}\right)}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}\frac{t^{-p+1}}{-p+1}-\frac{1^{-p+1}}{-p+1}}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}(\left(\frac{1}{1-p}\right)\frac{1}{t^{p-1}}-\frac{1}{1-p})}\\ & =& {\displaystyle 0-\frac{1}{1-p}=\frac{1}{p-1}\text{.}}\end{array}$$

Let $p=1\text{.}$ Evaluate ${\int }_1^{\infty }\frac{1}{x^p}dx$

$$\begin{array}{c} 1 x 1 y \end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_1^{\infty }\frac{1}{x^p}dx}& =& {\displaystyle {\int }_1^{\infty }\frac{1}{x}dx}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}{\int }_1^t{x}^{-1}dx}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}\left(\text{log}\hspace{0.17em}x{|}_{x=1}^{x=t}\right)}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}(\text{log}\hspace{0.17em}t-\text{log}\hspace{0.17em}1)}\\ & =& {\displaystyle \underset{t\to \infty }{\text{lim}}\text{log}\hspace{0.17em}t\text{diverges since}}\end{array}$$ $$\begin{array}{c} y=log t 1 t y \end{array}$$ Alternatively, $$\begin{array}{c} 1 2 3 1 1 2 \end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_1^{\infty }\frac{1}{x}dx}& >& {\displaystyle \frac{1}{2}+\underbrace{1/3+\frac{1}{4}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}+\cdots }\\ & >& {\displaystyle \frac{1}{2}+\frac{2}{4}+\frac{4}{8}+\cdots }\\ & =& {\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \text{diverges.}}\end{array}$$

Evaluate ${\int }_0^1\frac{1}{x^{\frac{1}{2}}}dx$

$$\begin{array}{c} y=1x12 1 x 1 y \end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_0^1{x}^{-\frac{1}{2}}dx}& =& {\displaystyle \underset{t\to 0}{\text{lim}}{\int }_1^t{x}^{-\frac{1}{2}}dx}\\ & =& {\displaystyle \underset{t\to 0}{\text{lim}}\left(2x^{\frac{1}{2}}{|}_{x=t}^{x=1}\right)}\\ & =& {\displaystyle \underset{t\to 0}{\text{lim}}(2-2t^{\frac{1}{2}})=2\text{.}}\end{array}$$

Evaluate ${\int }_{-1}^1\frac{1}{\sqrt{1-x^t}}dx$ $$\begin{array}{c} -1 1 x 1 y \end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{-1}^1\frac{1}{\sqrt{1-x^2}}dx}& =& {\displaystyle 2{\int }_0^1\frac{1}{\sqrt{1-x^2}}dx}\\ & =& {\displaystyle \underset{t\to 1}{\text{lim}}2{\int }_0^t\frac{1}{\sqrt{1-x^2}}dx}\\ & =& {\displaystyle \underset{t\to 1}{\text{lim}}\left(2\left({\text{sin}}^{-1}x\right){|}_{x=0}^{x=t}\right)}\\ & =& {\displaystyle \underset{t\to 1}{\text{lim}}(2{\text{sin}}^{-1}t-2{\text{sin}}^{-1}0)}\\ & =& {\displaystyle 2\frac{\pi }{2}-2·0=\pi \text{.}}\end{array}$$ $$\begin{array}{c} y=sin-1x -1 1 x - π 2 π 2 y \end{array}$$

Let $p\in {ℝ}_{\ge 0}$ with $p<1\text{.}$ Evaluate ${\int }_0^1\frac{1}{x^p}dx\text{.}$ (Think $p=\frac{1}{2}$ for comfort.) $$\begin{array}{ccc}{\displaystyle {\int }_0^1\frac{1}{x^p}dx}& =& {\displaystyle \underset{t\to 0}{\text{lim}}{\int }_1^t\frac{1}{x^p}dx}\\ & =& {\displaystyle \underset{t\to 0}{\text{lim}}\left(\frac{x^{-p+1}}{-p+1}{|}_{x=t}^{x=1}\right)}\\ & =& {\displaystyle \underset{t\to 0}{\text{lim}}(\frac{1^{-p+1}}{-p+1}-\frac{t^{-p+1}}{-p+1})}\\ & =& {\displaystyle \underset{t\to 0}{\text{lim}}(\frac{1}{1-p}-\frac{t^{1-p}}{1-p})}\\ & =& {\displaystyle \frac{1}{1-p}-0=\frac{1}{1-p}\text{.}}\end{array}$$

Evaluate ${\int }_0^1\frac{1}{x}dx$ $$\begin{array}{c} 1 x 1 y \end{array}$$ From the picture, $${\int }_0^1\frac{1}{x}dx=1+{\int }_1^{\infty }\frac{1}{x}dx\text{diverges (from earlier computation).}$$

Let $p\in ℝ$ with $p>1\text{.}$ Evaluate ${\int }_0^1\frac{1}{x^p}dx$ $$\begin{array}{c} y=1xp 1 x 1 y \end{array}$$ Between $x=0$ and $x=1,$ $\frac{1}{x^p}\ge \frac{1}{x}\text{.}$ So $${\int }_0^1\frac{1}{x^p}dx\ge {\int }_0^1\frac{1}{x}dx\text{which diverges.}$$ So $${\int }_0^1\frac{1}{x^p}dx\hspace{0.17em}$$ diverges.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100412Lect15.pdf and was given on 12 April 2010.

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