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

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?

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$\int \frac{1}{(x-1)^{1/3}} = \frac{3}{2}(x-1)^{2/3}$ – Stefan Hansen Jan 24 '13 at 19:05
Did you copy that correctly? The formulas with the limits make no sense, if you take a closer look. – Harald Hanche-Olsen Jan 24 '13 at 19:07
The idea is to integrate from $0$ to $u$, where $u\lt 1$, and from $v$ to $4$, where $v\gt 1$. Add up. Take the limit as $u$ approaches $1$ from the left, and $v$ approaches $1$ from the right. In each case, the antiderivative is $\frac{3}{2}(x+1)^{2/3}$. – André Nicolas Jan 24 '13 at 19:08

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.

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Thanks for all the help. I think there must be a mistake in the book as they've printed: $$\displaystyle\left.\lim_{y \to 1+}\frac{3}{2}(x-1)^{2/3}\right|_0^4$$ when, as is rightly stated here, it should be: $$\displaystyle\left.\lim_{y \to 1+}\frac{3}{2}(x-1)^{2/3}\right|_y^4$$ for the continuity from the right. I'll hopefully be able to look at this problem again tomorrow and if I need any more help, I'll hopefully be able to get some here. – Ian Jan 24 '13 at 19:42
You're welcome. – Stefan Hansen Jan 24 '13 at 19:43
Yes Ian, the first expression doesn't even involve $y$. – Stefan Hansen Jan 24 '13 at 20:20

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

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So; $1-\delta\to 1^{-}$ as $\delta\to 0$ and as Stefan noted? I want Ian to be sure of it. Thanks. – Babak S. Jan 24 '13 at 19:14
The identity in this post is wrong. See @Stefan's answer for a correct formulation. – Did Jan 26 '13 at 5:14
@Did: What's wrong with this post? – Mhenni Benghorbal Jan 26 '13 at 5:28
You know you do not care, so why do you ask? (Please read: The identity in this post is wrong. This tells you that the thing which is wrong in this post is the identity. And the identity is the line which has an $=$ in the middle.) – Did Jan 26 '13 at 9:16
@Did:Why are you assuming that I do not care? Thanks for the comment! – Mhenni Benghorbal Jan 26 '13 at 23:39

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