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

[jorgegalvan93]

Homework Statement

∫ a= 2 b = ∞ (dv)/(v^2+7v-8)

Homework Equations

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

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?

 

[SammyS]

Homework Statement

∫ a= 2 b = ∞ (dv)/(v^2+7v-8)

Homework Equations

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

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?

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.

 

[jorgegalvan93]

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.

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

 

[Dick]

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

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

 

[SammyS]

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

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

ln(a) - ln(b) ≠ ln(a)/ln(b) .

 

[Miliman13]

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

 

[Ray Vickson]

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

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?