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: \[\int_a^b f(x)\,\dd x= \lim_{v\downarrow a} \lim_{u\uparrow b} \int_u^v f(x)\,\dd x= \lim_{v\uparrow b} F(v) - \lim_{u\downarrow a} F(u).\]
\[\begin{aligned}\int_{0}^{2}\frac{1}{\sqrt{x}}\,\dd x&= \lim_{u\downarrow 0}\int_{u}^{2}\frac{1}{\sqrt{x}}\,\dd x\\ \\ &= \lim_{u\downarrow 0}\Bigl[2\sqrt{x}\Bigr]_u^{2}\\ \\ &= \lim_{u\downarrow 0}\left(2\sqrt{2} -2\sqrt{u}\right)\\ \\ &= 2\sqrt{2}\end{aligned} \]
\(\displaystyle \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}}\,\dd x = \pi\)
\[\begin{aligned} \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}}\,\dd x&= \lim_{u\downarrow -1}\lim_{v\uparrow 1} \int_u^{v} \frac{1}{\sqrt{1-x^2}}\,\dd x\\ \\ &= \lim_{u\downarrow -1} \lim_{v\uparrow 1} \bigl[\arcsin x \bigr]_{u}^{v}\\ \\ &= \lim_{v\uparrow 1}\arcsin v - \lim_{u\downarrow -1}\arcsin u\\ \\ &=\frac{\pi}{2}- (-\frac{\pi}{2}) = \pi\end{aligned}\]
