Is the Integral from -1 to 2 of 1/x Defined or Divergent?

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Discussion Overview

The discussion revolves around the evaluation of the definite integral $\displaystyle \int_{-1}^{2} \frac{d x}{x}$. Participants explore whether this integral is defined or divergent, considering various mathematical approaches and interpretations.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the integral is not defined due to the presence of a singularity at $x=0$, leading to divergent behavior when split into two parts.
  • Others propose that by symmetry, one might argue that $\int_{-1}^{0} \frac{dx}{x} + \int_{0}^{1} \frac{dx}{x} = 0$, although this is challenged as being non-rigorous.
  • A participant suggests that the Cauchy principal value could be considered, leading to a value of $\ln 2$ if evaluated correctly.
  • There is a discussion about the proper limits to take when evaluating the integral, with some emphasizing the need to consider limits simultaneously to avoid misinterpretation of divergent sums.
  • One participant presents two alternatives for interpreting the integral: treating it as undefined or assigning it a value of $\ln 2$ based on context, such as academic versus practical applications.
  • Another participant notes that unconventional definitions of convergence may apply, depending on the context in which the integral arises.
  • There is a mention of the Cauchy Principal Value Theorem, with some participants agreeing that it provides a way to assign a value to the integral despite its divergence.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the integral is defined or divergent. Multiple competing views remain regarding the interpretation and evaluation of the integral, particularly concerning the use of the Cauchy principal value.

Contextual Notes

Limitations in the discussion include the dependence on how limits are taken and the ambiguity surrounding the treatment of improper integrals. Participants express differing opinions on the appropriateness of assigning values to divergent integrals based on context.

chisigma
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In a site of Maths it has been requested to solve the following definite integral...

$\displaystyle \int_{-1}^{2} \frac{d x}{x} $ (1)

Well!... now I ask You all: what is the most appropriate answer?...

Kind regards

$\chi$ $\sigma$
 
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chisigma said:
In a site of Maths it has been requested to solve the following definite integral...

$\displaystyle \int_{-1}^{2} \frac{d x}{x} $ (1)

Well!... now I ask You all: what is the most appropriate answer?...

Kind regards

$\chi$ $\sigma$

$\displaystyle\int_{-1}^{2}\frac{dx}{x}$ is not defined. To see why, split the integral as

$\displaystyle\int_{-1}^{2}\frac{dx}{x}=\int_{-1}^0\frac{dx}{x}+\int_0^2\frac{dx}{x}$

$\displaystyle\int_{-1}^0\frac{dx}{x}=-\infty$ and $\displaystyle\int_0^2\frac{dx}{x}=+\infty$
 
I know this isn't rigorous at all, but intuitively couldn't one argue that by symmetry [math]\int_{-1}^{0} \frac{dx}{x}+\int_{0}^{1} \frac{dx}{x}=0[/math]?
 
Jameson said:
I know this isn't rigorous at all, but intuitively couldn't one argue that by symmetry [math]\int_{-1}^{0} \frac{dx}{x}+\int_{0}^{1} \frac{dx}{x}=0[/math]?

No, because $\displaystyle -\infty+\infty\neq 0$. (I see where you're coming from, though.)
 
Jameson said:
I know this isn't rigorous at all, but intuitively couldn't one argue that by symmetry [math]\int_{-1}^{0} \frac{dx}{x}+\int_{0}^{1} \frac{dx}{x}=0[/math]?
Be careful. Those are improper integrals. It's only zero if you evaluate the two limits at the same time, namely EDIT: $\displaystyle \lim_{\epsilon \to 0^{+}} \left( \int_{-1}^{-\epsilon} \frac{dx}{x} + \int_{\epsilon}^{1} \frac{dx}{x} \right) =0.$ Then what you have is the Cauchy principal value of a divergent integral.
 
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Random Variable said:
Be careful. Those are improper integrals. It's only zero if you evaluate the two limits at the same time, namely $\displaystyle \lim_{\epsilon \to 0} \left( \int_{-1}^{\epsilon} \frac{dx}{x} + \int_{\epsilon}^{1} \frac{dx}{x} \right) =0.$ Then what you have is the Cauchy principal value of a divergent integral.

This limit is more what I was thinking than just [math]\infty - \infty = 0[/math]. My reasoning was more along the lines of the above post. Random Variable - Shouldn't the first integral be [math]\int_{-1}^{-\epsilon} \frac{dx}{x}[/math] with the upper bound being negative epsilon?

I knew that my response was most likely incorrect but thought it would be a common reply so why not post it?
 
if someone ask me that i will draw
View attachment 121

and say it is the area above the x-axis minus below x-axis
 

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Jameson said:
This limit is more what I was thinking than just [math]\infty - \infty = 0[/math]. My reasoning was more along the lines of the above post. Random Variable - Shouldn't the first integral be [math]\int_{-1}^{-\epsilon} \frac{dx}{x}[/math] with the upper bound being negative epsilon?

Yes. And even more appropriately, it should should be $\displaystyle \lim_{\epsilon \to 0^{+}} \left( \int_{-1}^{-\epsilon} \frac{dx}{x} + \int_{\epsilon}^{1} \frac{dx}{x} \right)$.
 
Ok, that looks exactly how I was reasoning so it's good to know. As for the OP's question, how would you answer Random Variable? What is the most reasonable answer?
 
  • #10
Jameson said:
Ok, that looks exactly how I was reasoning so it's good to know. As for the OP's question, how would you answer Random Variable? What is the most reasonable answer?

It's a divergent integral because both $\displaystyle \lim_{\epsilon \to 0^{-}} \int_{-1}^{\epsilon} \frac{dx}{x}$ and $\displaystyle \lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{2} \frac{dx}{x} \ dx $ are not finite.

But if you want to assign it a value, it's Cauchy principal value is $\ln 2$.

If an integral does converge, BTW, it is equal to it's Cauchy principal value.
 
  • #11
Well!... it seems to me that we all agree on the fact that is...

$\displaystyle \int_{-1}^{2} \frac{d x}{x}= \int_{-1}^{1} \frac{d x}{x} + \int_{1}^{2} \frac{d x}{x}$ (1)

... so that the indefinite integral is the sum of two terms. The second term is $\displaystyle \int_{1}^{2} \frac{d x}{x}= \ln 2$ but what about the first term?... we have two alternatives...

a) we consider $\displaystyle \int_{-1}^{1} \frac{d x}{x}$ 'undefined' so that the whole integral is 'undefined'...

b) we consider $\displaystyle \int_{-1}^{1} \frac{d x}{x}=0$ so that the whole integral is $\displaystyle \ln 2$...

Both alternatives are 'reasonable' but what is the 'right alternative'?... my opinion is that the 'right alternative' strongly depends from the real contest. For example if the integral is part of a university test, then the the alternative a) is strongly recommended... but if the integral is part of an effective problem the solution of which involves millions of dollars then the alternative b) is strongly recommended...

Kind regards

$\chi$ $\sigma$
 
  • #12
chisigma said:
In a site of Maths it has been requested to solve the following definite integral...

$\displaystyle \int_{-1}^{2} \frac{d x}{x} $ (1)

Well!... now I ask You all: what is the most appropriate answer?...

Kind regards

$\chi$ $\sigma$

Obviously we all realize that this is undefined, but exactly the same difficulty does not stop us from summing divergent series when some other form of convergence than the usual is appropriate. But that leaves us with the problem of which unconventional definition of convergence we will use? I think the answer to that depends on context; how did this arise, what are we going to do with the answer?

CB
 
  • #13
Random Variable said:
Be careful. Those are improper integrals. It's only zero if you evaluate the two limits at the same time, namely $\displaystyle \lim_{\epsilon \to 0} \left( \int_{-1}^{\epsilon} \frac{dx}{x} + \int_{\epsilon}^{1} \frac{dx}{x} \right) =0.$ Then what you have is the Cauchy principal value of a divergent integral.

But it should be:

$\displaystyle \lim_{\epsilon, \delta \to 0} \left( \int_{-1}^{\epsilon} \frac{dx}{x} + \int_{\delta}^{1} \frac{dx}{x} \right) $

which does not exist.

CB
 
  • #14
CaptainBlack said:
But it should be:

$\displaystyle \lim_{\epsilon, \delta \to 0} \left( \int_{-1}^{\epsilon} \frac{dx}{x} + \int_{\delta}^{1} \frac{dx}{x} \right) $

which does not exist.

CB

Look at post #8.
 
  • #15
The appropriate answer in my complex analysis class is to use Cauchy's Principal Value Theorem and get $\ln 2$.
Similarly, for
$$
\int_0^3\frac{1}{x - 1}dx = \ln 2
$$
 
  • #16
\displaystyle \int_{-1}^{1} \frac{dx}{x} = 0 only if you take the limit in the way defined by the Cauchy principal value.

You could take the limit in another way and conclude that \displaystyle \int_{-1}^{1} \frac{dx}{x} = -\ln 2

http://en.wikipedia.org/wiki/Cauchy_principal_value#ExamplesFor a convergent integral, it shouldn't matter how you take the limit.

So I would conclude that it's never appropriate to say that \int_{-1}^{1} \frac{dx}{x} = 0. What is appropriate is $\displaystyle \text{PV} \int_{-1}^{1} \frac{dx}{x} = 0$.
 
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