The improper integral is defined as the limit of a convergent sequence of Riemann integrals for the following three cases. We also say that the improper integral converges (or is convergent).
Let $f:[a,\infty)$ be a real-valued function such that
Then we set the improper interval to this limit:
$$\int_a^\infty f(x)dx:=\lim_{b\to\infty}\int_a^bf(x)dx$$
Analogously, if $f:(\infty,b]$ is a real-valued function such that
then we set
$$\int_{-\infty}^b f(x)dx:=\lim_{a\to-\infty}\int_a^bf(x)dx.$$
If $f:[a,b)\to\mathbb R$ is a function such that
then we set
$$\int_a^b f(x)dx:=\lim_{\epsilon\searrow 0}\int_a^{b-\epsilon}f(x)dx.$$
Analogously, if $f:(a,b]\to\mathbb R$
then we set
$$\int_a^b f(x)dx:=\lim_{\epsilon\searrow 0}\int_{a+\epsilon}^bf(x)dx.$$
Let $f:(a,b)\to\mathbb R$ for $a\in\mathbb R\cup \{-\infty\}$ and $b\in\mathbb R\cup \{+\infty\}$ ($a,b$ being elements of the extended real numbers) is a function and let $c\in(a,b)$ be given^{1}. If
then we set
$$\int_a^b f(x)dx:=\lim_{\alpha\searrow a}\int_\alpha^{c}f(x)dx+\lim_{\beta\nearrow b}\int_c^{\beta}f(x)dx.$$
Proofs: 1 2 3
Propositions: 4 5
Note that this definition is independent of the specific choice of $c\in(a,b).$ This is because by hypothesis, $f$ is Riemann-integrable for all closed intervals $[\alpha,\beta]\subset (a,b)$ and the integrals can be summed to the same value on any adjacent intervals. ↩