Opening the book…
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
\[\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x=-\int_{+\infty}^{0} \frac{1}{1+u^{2}} d u=\int_{0}^{+\infty} \frac{1}{1+u^{2}} d u.\]
This page titled 7.7: An Improper Integral was last modified on Sun, 05 Sep 2021 15:44:32 GMT and is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform.
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