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The most appropriate answer...

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

Alexmahone

Active member
Jan 26, 2012
268
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 $\displaystyle\int_0^2\frac{dx}{x}=+\infty$
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
I know this isn't rigorous at all, but intuitively couldn't one argue that by symmetry \(\displaystyle \int_{-1}^{0} \frac{dx}{x}+\int_{0}^{1} \frac{dx}{x}=0\)?
 

Alexmahone

Active member
Jan 26, 2012
268
I know this isn't rigorous at all, but intuitively couldn't one argue that by symmetry \(\displaystyle \int_{-1}^{0} \frac{dx}{x}+\int_{0}^{1} \frac{dx}{x}=0\)?
No, because $\displaystyle -\infty+\infty\neq 0$. (I see where you're coming from, though.)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
I know this isn't rigorous at all, but intuitively couldn't one argue that by symmetry \(\displaystyle \int_{-1}^{0} \frac{dx}{x}+\int_{0}^{1} \frac{dx}{x}=0\)?
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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Jameson

Administrator
Staff member
Jan 26, 2012
4,041
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 \(\displaystyle \infty - \infty = 0\). 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?

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

Amer

Active member
Mar 1, 2012
275
if someone ask me that i will draw
Untitled.jpg

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

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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?
Yes. And even more appropriately, it should should be $\displaystyle \lim_{\epsilon \to 0^{+}} \left( \int_{-1}^{-\epsilon} \frac{dx}{x} + \int_{\epsilon}^{1} \frac{dx}{x} \right)$.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
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?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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.
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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 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?

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The appropriate answer in my complex analysis class is to use Cauchy's Principal Value Theorem and get $\ln 2$.
Similarly, for
$$
\int_0^3\frac{1}{x - 1}dx = \ln 2
$$
 

Random Variable

Well-known member
MHB Math Helper
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 $\displaystyle \text{PV} \int_{-1}^{1} \frac{dx}{x} = 0$.
 
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