7.7: An Improper Integral
- Page ID
- 25413
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)If \(f\) integrable on \([a, b]\) for all \(b>a\) and
\[\lim _{b \rightarrow+\infty} \int_{a}^{b} f(x) d x\]
exists, then we define
\[\int_{a}^{+\infty} f(x) d x=\lim _{b \rightarrow+\infty} \int_{a}^{b} f(x) d x.\]
If \(f\) is integrable on \([a, b]\) for all \(a<b\) and
\[\lim _{a \rightarrow-\infty} \int_{a}^{b} f(x) d x\]
exists, then we define
\[\int_{-\infty}^{b} f(x) d x=\lim _{a \rightarrow-\infty} \int_{a}^{b} f(x) d x.\]
Both of these integrals are examples of improper integrals.
Suppose \(f\) is continuous on \([a, \infty)\) and \(f(x) \geq 0\) for all \(x \geq a .\) If there exists \(g:[a,+\infty) \rightarrow \mathbb{R}\) for which
\[\int_{a}^{+\infty} g(x) d x\]
exists and \(g(x) \geq f(x)\) for all \(x \geq a\), then
\[\int_{a}^{+\infty} f(x) d x\]
exists.
Prove the preceding proposition.
Suppose
\[f(x)=\frac{1}{1+x^{2}}\]
and
\[g(x)=\left\{\begin{array}{ll}{1,} & {\text { if } 0 \leq x<1}, \\ {\frac{1}{x^{2}},} & {\text { if } x \geq 1}.\end{array}\right.\]
Then, for \(b>1\)
\[\int_{0}^{b} g(x) d x=\int_{0}^{1} d x+\int_{1}^{b} \frac{1}{x^{2}} d x=1+1-\frac{1}{b}=2-\frac{1}{b},\]
so
\[\int_{0}^{+\infty} g(x) d x=\lim _{b \rightarrow \infty}\left(2-\frac{1}{b}\right)=2.\]
Since \(0<f(x) \leq g(x)\) for all \(x \geq 0\), it follows that
\[\int_{0}^{+\infty} \frac{1}{1+x^{2}} d x\]
exists, and, moreover,
\[\int_{0}^{+\infty} \frac{1}{1+x^{2}} d x<2.\]
Also, the substitution \(u=-x\) shows that
\[\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x=-\int_{+\infty}^{0} \frac{1}{1+u^{2}} d u=\int_{0}^{+\infty} \frac{1}{1+u^{2}} d u.\]