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In these lecture notes, please see Example 5 on slide 11/15.

https://www2.math.upenn.edu/~ryblair/Math104/papers/Lec3_12Sol.pdf

This improper integral is given and we should study its convergence/divergence.

$$\int_{0}^{3}\frac{1}{(x-1)^\frac{2}{3}} dx$$

The notes say we should break it into two integrals, and study both of them separately. OK, that's fine (in some way) even though I was recently told here on this site that strictly speaking we don't have additivity (of this kind) for improper integrals. I mean, the equation $(1)$ below makes sense only if both integrals in the RHS are convergent. Anyway...

So we write:

$$\int_{0}^{3}\frac{1}{(x-1)^\frac{2}{3}} dx = \int_{0}^{1}\frac{1}{(x-1)^\frac{2}{3}} dx + \int_{1}^{3}\frac{1}{(x-1)^\frac{2}{3}} dx \tag{1}$$

and now we need to study both integrals on the RHS.

But now... I think then the first integral on the RHS is not well defined.

$$\int_{0}^{1}\frac{1}{(x-1)^\frac{2}{3}} dx \tag{2}$$

What does this last integral even mean? Should I treat it as:

$$\int_{0}^{1}\frac{1}{(1-x)^\frac{2}{3}} dx \tag{3}$$ (making the argument positive before taking it to power 2/3) because (in a way) it is essentially just

$$\int_{0}^{1}\frac{1}{\sqrt[3]{(x-1)^2}} dx \tag{4}$$

Or should I just say: OK, the expression $(x-1)^{2/3}$ is just undefined when $x \lt 1$

because when I type $$1 / ((x-1)^{\frac{2}{3}}) \tag{5}$$ in WA it says domain is $x \gt 1$.

But if it's undefined, how do I study the convergence/divergence of the first integral in $(1)$?!

So I don't think this example 5 is rigorous and well-defined.
Something doesn't make sense here, right?
What should this example even mean?

peter.petrov
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  • @DecarbonatedOdes Everything I am talking about here is in the context of real analysis. I see WA gives some answer involving complex numbers. But the lecturer given a different answer :) – peter.petrov Oct 22 '21 at 06:49
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    You can define a cubic root for all real numbers, see for example https://math.stackexchange.com/q/25528/42969. With that definition is $(x-1)^{2/3} = (\sqrt[3]{x-1})^2 = \sqrt[3]{(x-1)^2}$. – Martin R Oct 22 '21 at 06:55
  • Why is the domain $x\gt 1$? There should not be any problem even if $0\le x\lt 1$ considering that we are in realm of the real numbers. A square root instead of cube root would have created problem though. – Koro Oct 22 '21 at 06:56
  • Because in general the function $x^\alpha$ when $\alpha$ is any real number is defined only when $x \ge 0$ Otherwise $x^{1/3}$ and $x^{2/6}$ might be different things. – peter.petrov Oct 22 '21 at 06:57
  • But usually $(-1)^{\frac 1{2k+1}}:=-1$ for $k\in \mathbb Z$ (considering only the set of reals). So considering that, the integrand can also be defined, right? – Koro Oct 22 '21 at 07:00
  • @Koro As far as I know, it's not like that. Not in real analysis. Think about $x^{1/3}$ and $x^{2/6}$ See what happens when $x=-8$ for example. – peter.petrov Oct 22 '21 at 07:01
  • @Koro There's no cubic root here. The original problem/integral is written without cubic root. The expression $x^{2/3}$ is not quite the same thing as $\sqrt[3]{x^2}$ – peter.petrov Oct 22 '21 at 07:04
  • Related: What is $(-1)^{\frac{2}{3}}$? and How do you compute negative numbers to fractional powers?. – Martin R Oct 22 '21 at 07:11
  • @peter.petrov: I think that a meaning can be given to the expression (in $\mathbb R$) if we find $(-1)^{\frac 23}$ in the complex way $(re^{i\theta})$ and then pick the real value out of various possible (including complex values) values. – Koro Oct 22 '21 at 07:14
  • Clearly, the exercise assumes that $t^{2/3}$ is always defined and does equal $\sqrt[3]{t^2}$, also equal to $(\sqrt[3]t)^2$. –  Oct 22 '21 at 07:14
  • @peter.petrov: as $\frac13=\frac26$, $x^{1/3}$ and $x^{2/6}$ may not be defined differently. I can imagine three conventions for rational exponents: 1) undefined, 2) $x^{2/6}=x^{1/3}:=|x|^{1/3}$, 3) the fraction must be simplified before evaluation $x^{2/6}=x^{1/3}:=\sqrt[3]{x^2}$. –  Oct 22 '21 at 07:21
  • @YvesDaoust OK, I will solve this with the assumption $t^{2/3} = \sqrt[3]{t^2}$ and see what I will get. But my point was: would you (and others here) agree this improper integral is stated in a confusing, non-rigorous way? – peter.petrov Oct 22 '21 at 07:25
  • @peter.petrov: I cannot infer how the rational exponents were introduced to this student in the frame of this course. –  Oct 22 '21 at 07:26
  • @YvesDaoust Hm, I thought these definitions are universal and not course-specific. Strange... – peter.petrov Oct 22 '21 at 07:28
  • @peter.petrov: the OP refers to WA. But he forgot to read that the principal root is undefined for negative arguments, but the real-valued root is defined everywhere. I would be careful when talking about "universal definitions". –  Oct 22 '21 at 07:34

2 Answers2

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OK, based on all the notes/comments which I got, I decided to assume (for the purposes of this exercise) that the lecturer meant this integral.

$$\int_{0}^{1}\frac{1}{\sqrt[3]{(x-1)^2}} dx$$

when he wrote:

$$\int_{0}^{1}\frac{1}{(x-1)^\frac{2}{3}} dx$$

With this assumption in mind and after some calculations, I get the same answer as the one in the slides: both integrals on the RHS are convergent and their sum is $3(1 + \sqrt[3]{2})$

peter.petrov
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You can save yourself all the trouble by considering the convergence of the integral

$$I:=\int_{0}^{3}\frac{dx}{|x-1|^{\frac{2}{3}}}.$$

You show that $I$ is convergent. This implies that the function $g(x):=\frac{1}{|x-1|^{\frac{2}{3}}}$ is integrable,

and since $$f(x):=\frac{1}{(x-1)^{\frac{2}{3}}}\leq g(x)$$ for all $x\neq 1$ then the function $f$ is integrable on $[0,3]$, i.e., $\int_{0}^{3}f$ is convergent.

Medo
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  • You can't quite do it this way. This idea probably works but first you have to break it into two integrals anyway (from 0 to 1 and from 1 to 3). The integral $I$ has a finite value only if the two integrals on the RHS converge. Maybe one can apply apply your idea to both integrals in the RHS of $(1)$ – peter.petrov Oct 22 '21 at 08:06
  • yes, hence the phrase "you can show that $I$ is convergent". I left the easy details to you. – Medo Oct 22 '21 at 08:07
  • Also, by assuming that $t^{2/3} = \sqrt[3]{t^2}$ I essentially already did what you're saying here. – peter.petrov Oct 22 '21 at 08:07
  • Yeah, OK... I see. We're on the same page, I think. Thanks. – peter.petrov Oct 22 '21 at 08:08
  • If you read the nice comments here, you realize that $t^{2/3}\neq \sqrt[3]{t^2}$ when $t<0$. I am applying a comparison test for convergence. You are interpreting ( choosing) the integrand to be real on the interval $[0,1]$. – Medo Oct 22 '21 at 08:13
  • Yes, I understand your approach. This is a real analysis exercise so the integrand has to be real. So I don't see this as a limitation/restriction of some sort here. – peter.petrov Oct 22 '21 at 08:26