Discontinuous Integral

Question

For

$$\displaystyle\int_0^4\! \frac{dx}{(x-1)^{1/3}}$$

there is a discontinuity at $x=1$. In the book I'm studying, it says for the continuity from the right:

$$\displaystyle\left.\lim_{x \to 1+}\frac{3}{2}(x-1)^{2/3}\right|_0^4 = \lim_{x \to 1+}\frac{3}{2}[9^{1/3}-(u-1)^{2/3}-1]$$

I don't understand where the $(u-1)^{2/3}$ term comes from. Can anyone help?

Answer 1

Split up the integral around the point of discontinuity: $$ \int_0^4 (x-1)^{-1/3}\,\mathrm dx=\lim_{y\to 1^-}\int_0^y(x-1)^{-1/3}\,\mathrm dx+\lim_{y\to 1^+}\int_y^4(x-1)^{-1/3}\,\mathrm dx\\ =\lim_{y\to 1^-}\left[\frac{3}{2}(x-1)^{2/3}\right]_0^y+\lim_{y\to 1^+}\left[\frac{3}{2}(x-1)^{2/3}\right]_y^4 $$ Calculate the expressions within the brackets, and take the limit.

Answer 2

You need to split the interval of integration where the problem is and then consider the limits as $\delta$ and $\epsilon$ go to zero $$\displaystyle\int_0^4\! \frac{dx}{(x-1)^{1/3}} = \lim_{\delta \to 0^{+}}\displaystyle\int_0^{1-\delta}\! \frac{dx}{(x-1)^{1/3}}+\lim_{\epsilon \to 0^{+}}\displaystyle\int_{1+\epsilon}^4\! \frac{dx}{(x-1)^{1/3}} $$