Question

integral -1 to 1 of 1/x
(please see attached for an attempt)
[!] I keep getting infinity - infinity (an indeterminate form, which is inconclusive..)

Answer 1

can't do it

Answer 2

?!?

Answer 3

Why?

Answer 4

It is an improper integral.

Answer 5

You split it up

Answer 6

first of all we have to make sense out of this improper integral

Answer 7

Yes

Answer 8

I split it up into two integrals..

Answer 9

\[\lim_{l\to \infty}\int_{-1}^l\frac{dx}{x}\]

Answer 10

No! There is an absolute value

Answer 11

oh no, it's abs val

Answer 12

The entire argument of the natural logarithm has an absolute value.

Answer 13

I used the union integral property to split up into integrals.

Answer 14

but still this only has a Cauchy-Principle value

Answer 15

I think

Answer 16

anti derivative is the log
and as you go to
\[\lim_{l\to 0^{-}}\int_{-1}^l\frac{dx}{x}\] is more like it

Answer 17

You split it up into two integrals.

Answer 18

yes and neither converges

Answer 19

infinity - infinity (last line)

Answer 20

the anti derivative of \(\frac{1}{x}=\ln(x)\)

Answer 21

take the limit as \(x\to 0^+\) and you get \(-\infty\)

Answer 22

consider only \[\int_0^1\frac{dx}{x}\] which means
\[\lim_{l\to 0^+}\int_l^1\frac{dx}{x}\]

Answer 23

Ok then you will get:

ln(1) - ln(0)

In this case, you get infinity.

Answer 24

anti derivative is the log
you get
\[\log(1)-\lim_{l\to 0^+}\log(l)\]
so no integral, it is unbounded

Answer 25

Helper satellite73, did you see my attempt (opening post attachment)

Answer 26

and integral does not equal infinity
it is either a number or it does not exist

Answer 27

it is not "indeterminant form"
you have to compute the limits separately
neither exist

Answer 28

The integral does not converge, it has only a Cauchy principle value (CPV)
http://en.wikipedia.org/wiki/Cauchy_principal_value

Answer 29

But then where did I go wrong? Why am I getting infinity for one, but -infinity for another..

Answer 30

I did compute the limits seperately.

Answer 31

no problem , that is correct

Answer 32

there is nothing wrong with your work

Answer 33

But then why am I getting indeterminate forms..

Answer 34

you are right
neither integral exists
one goes to infinity, the other to minus infinity in the limit, so integral does not exist

Answer 35

Assume that I do not know that Cauchy's principle that you are talking about

Answer 36

I did, that's why a sent you a link on it

Answer 37

what do you mean by "indetermint form"?

Answer 38

But if you have indeterminate form, that is inconclusive! That suggests that further manipulation is required...

Answer 39

indeterminant form?

Answer 40

indeterminate form: infinity - infinity, for example

Answer 41

oh no!

Answer 42

i see the confusion hold on

Answer 43

suppose you have something like
\[\lim_{x\to \infty}\sqrt{x^2+1}-\sqrt{x^2+2x}\]
there you have something that looks like \(\infty-\infty\) or indeterminant form, and you have to do more work to get the limit
but this is entirely different

Answer 44

in this case you do not consider it as one integral to compute, but rather two separate integrals, one on the left and one on the right

Answer 45

so it is like saying
\(\lim_{x\to \infty}\sqrt{x^2+1}\) which is infinite, i.e. the limit does not exist
you do not compute it as the above example because the integrals are separate and the limit in each is taken separately

Answer 46

think about what you are trying to compute

\[\int_0^1\frac{dx}{x}\] is supposed to represent the area under the curve bounded by \(x=0\) and \(x=1\)
that area is just too damned large

Answer 47

ok

Answer 48

so in short, your work is completely right, but the answer is not to be computed together as one limit, i.e. not computed as an indeterminant form
as soon as one integral does not have a finite limit, you are done

Answer 49

\[\large \mathrm{PV} \int_{-1}^{1} \frac{\mathrm{d}x}{x}=\lim_{\epsilon \rightarrow 0^+}\left [ \int_{-1}^{-\epsilon} \frac{\mathrm{d}x}{x} +\int_{\epsilon}^{1} \frac{\mathrm{d}x}{x}\right ] =0\]