Integrating improper integrals constitute of integrating functions
1) over an infinite integral
2) over an interval where f has a discontinuity.
Namely, integrals type I and type II, respectively. Generally, both types are solved in the same way using limits.
But consider the following integral:
$\int_0^\infty \frac{1}{\sqrt[3]{x}} dx$
In this case, I would proceed to set the integral $\int_0^t \frac{1}{\sqrt[3]{x}} dx$ to $\displaystyle\lim_{t\to \infty}$. However, $\frac{1}{\sqrt[3]{x}}$ is discontinuous at $x=0$.
Why is this disregarded during integration??
i think in that case you divide the integral in two because it have two cases of improper integral, so you make
$$\int_{0}^{+\infty}\frac{dx}{\sqrt[3]{x}}=\int_{0}^{1}\frac{dx}{\sqrt[3]{x}}+\int_{1}^{+\infty}\frac{dx}{\sqrt[3]{x}}$$ so you study both part separated and in order to the original integral converge both integrals should converge. so it will be something like
$$\int_{0}^{1}\frac{dx}{\sqrt[3]{x}}=\lim_{t\to0^+}\int_{t}^{1}\frac{dx}{\sqrt[3]{x}}$$
and
$$\int_{1}^{\infty}\frac{dx}{\sqrt[3]{x}}=\lim_{t\to\infty}\int_{1}^{t}\frac{dx}{\sqrt[3]{x}}$$