There are two types of improper integrals:

1) Type 1

a) $\int_{a}^{\infty}f(x) dx$ $=$ $\lim$

_{t →$\infty$}$\int_{a}^{t}f(x)dx$

b) $\int_{-\infty}^{b}f(x)dx=$$\lim$

_{t →$-\infty$}$\int_{t}^{b}f(x)dx$

c) $\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx$

2) Type 2

a) If $f$ is continuous on $[a,b)$ and discontinuous at $b$, then:

$\int_{a}^{b} f(x)dx=$$\lim$

_{t →$b^-$}$\int_{a}^{t}f(x)dx$

b) If $f$ is continuous on $(a,b]$ and discontinuous at $a$, then:

$\int_{a}^{b} f(x)dx=$$\lim$

_{t →$a^+$}$\int_{t}^{b}f(x)dx$

c) If $f$ has a discontinuity at $c$, where $a<c<b$, then:

$\int_{a}^{b} f(x)dx=\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx$

If the limits exist and is finite, then it is convergent. Otherwise, it is divergent.