
# 7.7: An Improper Integral


## 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 $$b>1$$

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

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

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