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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

$\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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- Thread starter
- #1

- Feb 13, 2012

- 1,704

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

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

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 268

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

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

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

Kind regards

$\chi$ $\sigma$

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

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

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- Jan 26, 2012

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- Jan 26, 2012

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No, because $\displaystyle -\infty+\infty\neq 0$. (I see where you're coming from, though.)

- Jan 31, 2012

- 253

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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- #6

- Jan 26, 2012

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

I knew that my response was most likely incorrect but thought it would be a common reply so why not post it?

- Jan 31, 2012

- 253

Yes. And even more appropriately, it should should be $\displaystyle \lim_{\epsilon \to 0^{+}} \left( \int_{-1}^{-\epsilon} \frac{dx}{x} + \int_{\epsilon}^{1} \frac{dx}{x} \right)$.This limit is more what I was thinking than just \(\displaystyle \infty - \infty = 0\). My reasoning was more along the lines of the above post. Random Variable - Shouldn't the first integral be \(\displaystyle \int_{-1}^{-\epsilon} \frac{dx}{x}\) with the upper bound being negative epsilon?

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- #9

- Jan 26, 2012

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- Jan 31, 2012

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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.

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- #11

- Feb 13, 2012

- 1,704

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

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

Obviously we all realise 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?

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

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

Kind regards

$\chi$ $\sigma$

CB

- Jan 26, 2012

- 890

But it should be: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.

$\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

- Jan 31, 2012

- 253

Look at post #8.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

- Jan 31, 2012

- 253

[tex]\displaystyle \int_{-1}^{1} \frac{dx}{x} = 0[/tex] 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 [tex]\displaystyle \int_{-1}^{1} \frac{dx}{x} = -\ln 2[/tex]

http://en.wikipedia.org/wiki/Cauchy_principal_value#Examples

For a convergent integral, it shouldn't matter how you take the limit.

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

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

http://en.wikipedia.org/wiki/Cauchy_principal_value#Examples

For a convergent integral, it shouldn't matter how you take the limit.

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

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