Question

Improper integral

Answer 1

Do I have to split this improper integral in 2 to solve it?
\[\int\limits_{0}^{1}\frac{ dx }{ x^2-1 }\]

Answer 2

ex at 1/2?

Answer 3

\[\lim_{y \rightarrow 1^+}\int\limits\limits_{0}^{y}\frac{ dx }{ x^2-1 }\]

Answer 4

take integral by partial fraction, to me.

Answer 5

\[\left[ \frac{ \ln(x-1) }{ 2 } - \frac{ \ln(x+1) }{ 2 } \right]_{0}^{y}\]

Answer 6

Absolute value for terms after ln. plug values of limits in ,then plug all them after lim to caclculate

Answer 7

\[\lim_{y \rightarrow 1^+}\left[ \frac{ \ln(y-1) }{ 2 } - \frac{ \ln(y+1) }{ 2 } \right]-\left[ \frac{ \ln(0-1) }{ 2 } - \frac{ \ln(0+1) }{ 2 } \right]\]

Answer 8

absolute value, please. ln|y-1| , take 1/2 out first, and then plug y =1 into there to take lim. that's it

Answer 9

ln1 =0, right?

Answer 10

yes

Answer 11

ok, go ahead, just one more step. you get the answer.

Answer 12

got what I mean?

Answer 13

Yes am with you.. the result is -∞

Answer 14

how? the result in book? oops!!

Answer 15

2 sec.

Answer 16

This is an upper limit now, so understand why the limit is actually from the left this time and not from the right :)
\[\Large \lim_{y \rightarrow \color{red}{1^-}}\int\limits\limits_{0}^{y}\frac{ dx }{ x^2-1 }\]

Answer 17

go ahead, @terenzreignz , please

Answer 18

I was just pointing out a minor error... I'm not even sure it affects much...
didn't you two already finish? It seemed like it :P

Answer 19

I don't know, to me, it is = -ln 2/2 ,but he said that it is = - infinite. eeeehhh.

Answer 20

Okay, let's do this again :D
\[\large \int\limits_{0}^{1}\frac{ dx }{ x^2-1 }\]

Answer 21

Splitting it into partial fractions...
\[\large \frac1{x^2-1}= \frac12\left[\frac{1}{x-1}-\frac{1}{x+1}\right]\]

Answer 22

So we integrate this instead...
\[\Large\int\limits_0^1 \frac{dx}{x^2-1}= \frac12\int\limits_0^1\left[\frac{1}{x-1}-\frac{1}{x+1}\right]dx\]

Answer 23

We get...
\[\Large \frac12\left[\ln(x-1)-\ln(x+1)\right]_0^1\]

Answer 24

Or rather...
\[\Large \lim_{b\rightarrow1^-}\frac12\left[\ln(x-1)-\ln(x+1)\right]_0^b\]

Answer 25

Everything fine up to here?
Please check. I'm oh-so prone to errors :3

Answer 26

It looks great.

Answer 27

so far so good, friend

Answer 28

Wait, I forgot absolute value.
...
Let's not forget that
\[\Large \int \frac1xdx = \ln|x|+C\]

Answer 29

\[\Large \lim_{b\rightarrow1^-}\frac12\left[\ln|x-1|-\ln|x+1|\right]_0^b\]
Much better.

Answer 30

It does... otherwise, we'd get ln of a negative 1.
\[\Large \lim_{b \rightarrow 1^-}\left[ \frac{ \ln(y-1) }{ 2 } - \frac{ \ln(y+1) }{ 2 } \right]-\left[ \frac{ \color{red}{\ln(0-1)} }{ 2 } - \frac{ \ln(0+1) }{ 2 } \right]\]

Answer 31

\[\Large \lim_{b \rightarrow 1^-}\left[ \frac{ \ln|b-1| }{ 2 } - \frac{ \ln|b+1| }{ 2 } \right]-\left[ \frac{ \color{red}{\ln|0-1|} }{ 2 } - \frac{ \ln|0+1| }{ 2 } \right]\]

Answer 32

yes, you are right for the second part. got you

Answer 33

So... Let's simplify.
\[\Large \lim_{b \rightarrow 1^-}\left[ \frac{ \ln|b-1| }{ 2 } - \frac{ \ln|b+1| }{ 2 } \right]-\color{red}{\left[ \frac{{\ln(1)} }{ 2 } - \frac{ \ln(1) }{ 2 } \right]}\]
and the part in red just becomes zero.

Answer 34

yup

Answer 35

\[\Large \lim_{b \rightarrow 1^-}\left[ \frac{ \ln|b-1| }{ 2 } - \frac{ \ln|b+1| }{ 2 } \right]\]
So everything's simpler now.

Answer 36

\[\Large \lim_{b \rightarrow 1^-}\left[ \color{blue}{\frac{ \ln|b-1| }{ 2 }} - \color{green}{\frac{ \ln|b+1| }{ 2 }} \right]\]

Blue part goes to negative infinity, green part goes to some constant \(\Large \frac{\ln(2)}
{2}\)
So the entire thingy goes to minus infinity :D

Answer 37

Everything fine now? :D

Answer 38

Still no? :)
This is a rough sketch of what ln looks like...

Answer 39

oh yea, because lim ln 0 = - infinitive.