Infinite areas cancel out?

**mrmidget81** · August 31st 2009, 08:06 PM

My math teacher and I got in a debate the other day, and i couldn't find any information anywhere so i figured i'd post it here

We were working on calculating infinite converging areas and we got to the following problem

$\begin{array}{cc}-1\\1\end{array}\int\frac{1}{x}dx$

Since i'm always looking for ways to avoid doing extra work, i said it would be possible to simply cancel out the areas from -1 to 0 and from 0 to 1 based on symetry and end up with zero

however, when you actually work out the integral you get $\infty$ - $\infty$ which is undefined

As far as i can tell, the reason that is undefined is because the two infinities could be coming from anywhere and it's impossible to make an assumption, but in this case, when we know the two infinite areas are exactly equal (except for the minus sign) why can't we say that is equal to zero?

**ynj** · August 31st 2009, 08:17 PM

Because for the positive part, say the "area" between [0,1], there area is actuall infinite, that is to say, does not exist. So you can not add the area between [-1,0] and [0,1] and say it is 0, since there is no "area" between [0,1].
Or you can explain like this:
This improper integral that have a spot 0 is defined iff $\int_0^1 f(x) \ dx,\int_{-1}^0 f(x) \ dx$ are both defined.

**mr fantastic** · September 1st 2009, 01:03 AM

Originally Posted by mrmidget81

My math teacher and I got in a debate the other day, and i couldn't find any information anywhere so i figured i'd post it here

We were working on calculating infinite converging areas and we got to the following problem

$\begin{array}{cc}-1\\1\end{array}\int\frac{1}{x}dx$

Since i'm always looking for ways to avoid doing extra work, i said it would be possible to simply cancel out the areas from -1 to 0 and from 0 to 1 based on symetry and end up with zero

however, when you actually work out the integral you get $\infty$ - $\infty$ which is undefined

As far as i can tell, the reason that is undefined is because the two infinities could be coming from anywhere and it's impossible to make an assumption, but in this case, when we know the two infinite areas are exactly equal (except for the minus sign) why can't we say that is equal to zero?

To add a little more from the previous post:

$\int_{-1}^{1} \frac{1}{x} \, dx$ is an improper integral. It is undefined and not necessarily well defined.

It can be taken to mean $\lim_{a \rightarrow 0} \left( \int_{-1}^{a} \frac{1}{x} \, dx + \int_{a}^{1} \frac{1}{x} \, dx \right) = 0$ , which is the Cauchy Principle Value.

But it can also be taken to mean $\lim_{a \rightarrow 0} \left( \int_{-1}^{2a} \frac{1}{x} \, dx + \int_{a}^{1} \frac{1}{x} \, dx \right) = \ln 2$ .... etc.

The previous poster has given the conditions under which an improper integral of this type will be convergent.

**mrmidget81** · September 1st 2009, 06:04 PM

Thanks for the quick replies

and sorry to ask a follow up question but how is it then you can calculate two different areas for the same integral?

**mr fantastic** · September 1st 2009, 07:28 PM

Originally Posted by mrmidget81

Thanks for the quick replies

and sorry to ask a follow up question but how is it then you can calculate two different areas for the same integral?

When things don't exist, contradictions and inconsistencies are often present ....

**mrmidget81** · September 1st 2009, 08:16 PM

but can you really say it doesn't exist just because you can't define it?

**ynj** · September 1st 2009, 08:32 PM

Originally Posted by mrmidget81

but can you really say it doesn't exist just because you can't define it?

The area is defined by integral...
So when integral does not exists, area does not exists..

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