1. ## The Definite Integral

So, i have a problem that i put into wolfram and maple and cant replicate the output..
What am i doing wrong?

$\int_2^{\infty} \frac{2}{v^2 - v} \,dv$

Then,

$\frac{2}{v^2 -v} = \frac {-2}{v} + \frac{2}{v - 1}$

$-2 \int_2^{\infty} \frac{1}{v}\,dv + 2 \int_2^{\infty} \frac{1}{v - 1}\,dv$

So, i would have leaving out absolute

$-2 \ln v + 2 \ln v -1 ]_2^{\infty}$

evaluating i have

$-2 \ln {\infty} + 2 \ln {\infty} -1 + 2 \ln 2 + 2 \ln 2 -1$

so, the left side falls out to zero and the right side is 4 ln4

However,

wolfram solution ln4
maple solution: 2 ln 2

Which look equivalent.

What am i doing so wrong?

2. ## Re: The Definite Integral

Originally Posted by icelated
$-2 \int_2^{\infty} \frac{1}{v}\,dv + 2 \int_2^{\infty} \frac{1}{v - 1}\,dv$
So, i would have leaving out absolute
$-2 \ln v + 2 \ln v -1 ]_2^{\infty}$
Note that $-2 \ln v + 2 \ln v -1 ]=2\ln \left( {\frac{{v - 1}}{v}} \right) = 2\ln \left( {1 - \frac{1}{v}} \right)_2^\infty$

3. ## Re: The Definite Integral

How did you get v =

Originally Posted by Plato
Note that $=2\ln \left( {\frac{{v - 1}}{v}} \right) = 2\ln \left( {1 - \frac{1}{v}} \right)_2^\infty$
How did you get that fraction, that makes no sense to me!

4. ## Re: The Definite Integral

Originally Posted by icelated
How did you get v =
How did you get that fraction, that makes no sense to me!
That is a basic rule of logarithm.
$-\ln(a)+\ln(b)=\ln \left( {\frac{b}{a}} \right)$

$\lim _{b \to \infty } \left. {2\ln \left( {1 - \frac{1}{v}} \right)} \right|_2^b = \lim _{b \to \infty } 2\ln \left( {1 - \frac{1}{b}} \right) - 2\ln \left( {1 - \frac{1}{2}} \right) = - 2\ln \left( {\frac{1}{2}} \right)$

5. ## Re: The Definite Integral

Of course. The things we forget! Or some of us.. Didnt even think about the log rule.
Thank you so much! You are very helpful....

6. ## Re: The Definite Integral

Originally Posted by icelated
Of course. The things we forget! Or some of us.. Didnt even think about the log rule.
Thank you so much! You are very helpful....
Bottom of every post is a thanks button, use it. There is no need to make a thanks post.

CB

7. ## Re: The Definite Integral

Originally Posted by icelated
$-2 \ln {\infty} + 2 \ln {\infty} -1 + 2 \ln 2 + 2 \ln 2 -1$
It is mathematically incorrect to substitute \displaystyle \begin{align*} \infty \end{align*} as a number.

For improper integrals, if \displaystyle \begin{align*} \int{f(x)\,dx} = F(x) + C \end{align*}, then...

\displaystyle \begin{align*} \int_a^{\infty}{f(x)\,dx} &= \lim_{\epsilon \to \infty}\int_a^{\epsilon}{f(x)\,dx} \\ &= \lim_{\epsilon \to \infty}F(\epsilon) - F(a)\end{align*}

8. ## Re: The Definite Integral

Originally Posted by icelated
So, i have a problem that i put into wolfram and maple and cant replicate the output..
What am i doing wrong?

$\int_2^{\infty} \frac{2}{v^2 - v} \,dv$

Then,

$\frac{2}{v^2 -v} = \frac {-2}{v} + \frac{2}{v - 1}$

$-2 \int_2^{\infty} \frac{1}{v}\,dv + 2 \int_2^{\infty} \frac{1}{v - 1}\,dv$

So, i would have leaving out absolute
which you can do because 2> 1 so x-1> 0 for all x> 2

$-2 \ln v + 2 \ln v -1 ]_2^{\infty}$

evaluating i have

$-2 \ln {\infty} + 2 \ln {\infty} -1 + 2 \ln 2 + 2 \ln 2 -1$
Your error is leaving out parentheses and fooling yourself!
$-2\ln(\infty)+ 2\ln(\inty- 1) + 2\ln(2)+ 2\ln(2- 1)$
The first two terms cancel leaving $2ln(2)+ 2\ln(1)= 2\ln(2)$.

so the left side falls out to zero and the right side is 4 ln4

However,

wolfram solution ln4
maple solution: 2 ln 2

Which look equivalent.

What am i doing so wrong?