Improper integrals of type 2

Improper integrals: Oneigenlijke integralen *

Improper integrals of type 2

Let $f(x)$ be a given integrand, and $F(x)$ an antiderivative of $f(x)$ . We consider three cases of integration domains where $f(x)$ s unbounded near at least one of the boundary points.

If $f(x)$ is continuous on the interval $[a,b)$ and if $\displaystyle \lim_{v\uparrow b} F(v)$ exists, then:

$\int_a^b f(x)\,\dd x = \lim_{v\uparrow b} \int_a^{v} f(x)\,\dd x= \lim_{v\uparrow b} F(v) - F(a)$

If $f(x)$ is continuous on the interval $(a,b]$ and if $\displaystyle \lim_{u\downarrow a} F(u)$ exists, then:

$\int_a^b f(x)\,\dd x= \lim_{u\downarrow a} \int_{u}^b f(x)\,\dd x= F(b) - \lim_{u\downarrow a} F(u)$

If $f(x)$ is continuous on the interval $(a,b)$ and if the two limits $\displaystyle \lim_{v\uparrow b} F(v)$ en $\displaystyle \lim_{u\downarrow a} F(u)$ exist and if at least one of them is finite, then:

$\int_a^b f(x)\,\dd x= \lim_{v\downarrow a} \lim_{u\uparrow b} \int_u^v f(x)\,\dd x= \lim_{v\uparrow b} F(v) - \lim_{u\downarrow a} F(u).$

Calculate the following improper integral or show that it diverges.

$\int_{0}^{2}\frac{1}{\sqrt{x}}\,\dd x$

$\displaystyle\int_{0}^{2}\frac{1}{\sqrt{x}}\,\dd x={}$ $2\sqrt{2}$

$\begin{aligned}\int_{0}^{2}\frac{1}{\sqrt{x}}\,\dd x&= \lim_{u\downarrow 0}\int_{u}^{2}\frac{1}{\sqrt{x}}\,\dd x\\ \\ &= \lim_{u\downarrow 0}\Bigl[2\sqrt{x}\Bigr]_u^{2}\\ \\ &= \lim_{u\downarrow 0}\left(2\sqrt{2} -2\sqrt{u}\right)\\ \\ &= 2\sqrt{2}\end{aligned}$

New example

Calculate the following improper integral or show that it diverges.

$\int_{-1}^{1} \frac{1}{\sqrt{1-x^2}}\,\dd x$

Solution
$\displaystyle \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}}\,\dd x = \pi$

$\begin{aligned} \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}}\,\dd x&= \lim_{u\downarrow -1}\lim_{v\uparrow 1} \int_u^{v} \frac{1}{\sqrt{1-x^2}}\,\dd x\\ \\ &= \lim_{u\downarrow -1} \lim_{v\uparrow 1} \bigl[\arcsin x \bigr]_{u}^{v}\\ \\ &= \lim_{v\uparrow 1}\arcsin v - \lim_{u\downarrow -1}\arcsin u\\ \\ &=\frac{\pi}{2}- (-\frac{\pi}{2}) = \pi\end{aligned}$