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I am having some trouble with a problem very similar to this in my study guide, how can I start, the $-64$ is really intimidating to me.

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    $\begingroup$ What is the antiderivative of $x^k$ ? $\endgroup$ – Claude Leibovici Aug 16 '14 at 4:35
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    $\begingroup$ The integral is an improper integral, since our function blows up at $0$. But this improper integral converges. $\endgroup$ – André Nicolas Aug 16 '14 at 4:38
  • $\begingroup$ @AndréNicolas. Could you interfere in this no end discussion ? Thanks. $\endgroup$ – Claude Leibovici Aug 16 '14 at 5:51
  • $\begingroup$ What intervention do you suggest? The PV discussion is of no relevance because of the convergence. $\endgroup$ – André Nicolas Aug 16 '14 at 6:10
  • $\begingroup$ @Brian. May I ask you a favor ? When you will get the "official" answer from your professor, could you add it to your post. I am really curious. Thanks and cheers :-) $\endgroup$ – Claude Leibovici Aug 16 '14 at 7:23
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We have to break this up and take the limit.

$$\lim_{\epsilon\to 0} \left (\int_{-64}^{-\epsilon}x^{-\frac{1}{3}}dx+\int^{1}_{\epsilon}x^{-\frac{1}{3}}dx \right )$$

Using the power rule,

$$\lim_{\epsilon\to 0} \left (\left[\frac{3}{2}x^{\frac{2}{3}} \right ]_{-64}^{-\epsilon}+\left [\frac{3}{2}x^{\frac{2}{3}} \right ]^{1}_{\epsilon}\right )$$

This becomes

$$\lim_{\epsilon\to 0} \left (\frac{3}{2}(-\epsilon)^{\frac{2}{3}}-\frac{3}{2}(64)^{\frac{2}{3}}+\frac{3}{2}1^{\frac{2}{3}} -\frac{3}{2}\epsilon^{\frac{2}{3}} \right )$$ Notice the cancellation of the $\epsilon$ terms. So we are left with

$$=-\frac{3}{2}(4)^2+\frac{3}{2}=\frac{3}{2}(1-16)=-\frac{45}{2} $$

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  • $\begingroup$ This is not correct, I am afraid. $\endgroup$ – Claude Leibovici Aug 16 '14 at 5:05
  • $\begingroup$ @ClaudeLeibovici I agree with you. This is Cauchy principal value. $\endgroup$ – Cortizol Aug 16 '14 at 6:00
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There's really nothing to be intimidated of. Once you realize that $1/x^{1/3} = x^{-1/3}$, you can just use power rule.

Edit: I just realized that you probably want to take the riemann sum and compute it manually. In that case, I would suggest integrating the inverse.

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  • $\begingroup$ then would i just integrate? $\endgroup$ – brian Aug 16 '14 at 4:36
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    $\begingroup$ The first step is to get the antiderivative. Then use bounds. $\endgroup$ – Claude Leibovici Aug 16 '14 at 4:38
  • $\begingroup$ okay i got (3/2)-24(-1)^(2/3) thank you $\endgroup$ – brian Aug 16 '14 at 4:39
  • $\begingroup$ could that be simplified in any way $\endgroup$ – brian Aug 16 '14 at 4:40
  • $\begingroup$ uh.. (-1)^2/3 is 1. So you should get 3/2 - 24 = -45/2. $\endgroup$ – Tae Hyung Kim Aug 16 '14 at 4:47
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It is an improper integral . one may write $\int_{-64}^0x^\frac{-1}{3}dx+\int_{0}^1x^\frac{-1}{3}dx.$

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Hint

First, the antiderivative $$\int\frac{dx}{x^{\frac{1}{3}}}=\frac{3 x^{2/3}}{2}$$ Now, for the integral (taking into account André Nicolas's comment which does not need to be repeated), the result is then $$\frac{3}{2}\Big(1^{\frac{2}{3}}-(-64)^{\frac{2}{3}}\Big)=\frac{3}{2}\Big(1-(-64)^{\frac{2}{3}}\Big)$$ You were right to be "intimidated" by the $-64$ since $(-64)^{\frac{2}{3}}$ is a complex number which you need to simplify.

I suppose that this is the key part of the problem but I am sure that you can take from here.

Added later to this answer

I can accept to be totally wrong but, at least for me $$(-64)^{\frac{2}{3}}=-8+8 i \sqrt{3}$$

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  • $\begingroup$ I don't see how $(-64)^{\frac{2}{3}}$ is a complex number $\endgroup$ – Learning Aug 16 '14 at 5:14
  • $\begingroup$ @Mr.T. OK, what is its value ? $\endgroup$ – Claude Leibovici Aug 16 '14 at 5:15
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    $\begingroup$ Is it not true that $(-64)^{\frac{2}{3}}$ is $16$ in this situation? It's the integral over the real number set, and it's well-defined. Of course the structure $(-64)^{\frac{2}{3}}$ may be referred to a complex number, but this is not the case I think. As I know, $(-4)^3=-64$ and $x^{\frac{1}{3}}$ in this case is the third root of a real number. No complex needed $\endgroup$ – Learning Aug 16 '14 at 5:40
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    $\begingroup$ @JyrkiLahtonen. Could you clarify ? I really do not understand what you mean. And, almost, don't worry : I take easy ! Cheers :-) $\endgroup$ – Claude Leibovici Aug 16 '14 at 5:53
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    $\begingroup$ By the way I'm not the downvoter but I suspect it's your mention of complex numbers that was the reason, not the continuity issue that Jyrki mentioned, which you also should have mentioned. The point is that the complex logarithm can be defined with different branches, and in different situations we need to use a different branch. Your value for the cube-root is just one of the three possible values. It happens that we can't have a branch of the complex logarithm such that the cube-root function agrees with the conventional real-valued cube-root, but we should recognize which is being used. $\endgroup$ – user21820 Aug 16 '14 at 6:13

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