Determine whether the integral is divergent. If convergent, evaluate.

1. Feb 26, 2013

jorgegalvan93

1. The problem statement, all variables and given/known data
∫ a= 2 b = ∞ (dv)/(v^2+7v-8)

2. Relevant equations

I have attempted the problem and am confused as to why the integral is not divergent.

3. The attempt at a solution

I integrated the function by using partial fractions and came up with a result of:
-1/9ln(v+8)+1/9ln(v-1)

I replaced 'b' limit of integration with 't' and solved for the limit of the function as 't' approaches infinity…
lim t→∞ -1/9[ln(v+8)-ln(v-1)] with limits of integration, b =t and a = 2

However when finding the limit, I realize that when substituting ∞ for 'v' I am left with the following result: -1/9[ln(∞+8)-ln(∞-1)]

ln(∞) is equal to ∞, and ∞-∞ is equal to ∞. Therefore there is no limit for the function ∫ a= 2 b = ∞ (dv)/(v^2+7v-8) and it is divergent.

This is not the case though, and the function convergent.

Where is my mistake occurring?

2. Feb 26, 2013

SammyS

Staff Emeritus
Re: Determine whether the integral is divergent. If convergent, evalua

A limit of the form ∞ - ∞ is indeterminate.

You need to express it in a different way. use properties of logs to to get a more compact expression.

3. Feb 26, 2013

jorgegalvan93

Re: Determine whether the integral is divergent. If convergent, evalua

ln(v+8)/ln(v+1) ?
Would I then have to use L'hopitals rule?

4. Feb 26, 2013

Dick

Re: Determine whether the integral is divergent. If convergent, evalua

That's not a correct use of the rules of logs. Try again.

5. Feb 26, 2013

SammyS

Staff Emeritus
Re: Determine whether the integral is divergent. If convergent, evalua

ln(a) - ln(b) = ln(a/b) .

ln(a) - ln(b) ln(a)/ln(b) .

6. Oct 16, 2017

Miliman13

Can some one help me figure out How ln( t+8)/(t-1) = 0

Last edited: Oct 16, 2017
7. Oct 16, 2017

Ray Vickson

If you mean
$$\frac{\ln (t+8)}{t-1}=0$$
it is easy (figure out why). If you mean
$$\ln \left( \frac{t+8}{t-1} \right) = 0$$
you can spend a lifetime looking for a solution and never finding one. Why not?

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