7.7: An Improper Integral

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Definition

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.

Proposition \(\PageIndex{1}\)

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.

Exercise \(\PageIndex{1}\)

Prove the preceding proposition.

Example \(\PageIndex{1}\)

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 \(b>1\)

\[\int_{0}^{b} g(x) d x=\int_{0}^{1} d x+\int_{1}^{b} \frac{1}{x^{2}} d x=1+1-\frac{1}{b}=2-\frac{1}{b},\]

\[\int_{0}^{+\infty} g(x) d x=\lim _{b \rightarrow \infty}\left(2-\frac{1}{b}\right)=2.\]

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

\[\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x=-\int_{+\infty}^{0} \frac{1}{1+u^{2}} d u=\int_{0}^{+\infty} \frac{1}{1+u^{2}} d u.\]