Integral from 0 to infinity of: 1/[z^2 + 3z + 2] dz

[MarkSA]

Hello,

Determine whether the integral is divergent or convergent and evaluate it.

1) Integral from 0 to infinity of: 1/[z^2 + 3z + 2] dz

To solve, I did the limit as t->infinity of: integral from 0 to t of: 1/(z + 2)(z + 1) dz
It looks like partial fraction decomposition to me...

For partial fractions I end up with A/(z + 2) and B/(z + 1), with A = -1 and B = 1
This gets me:

limit as t-> infinity of: [(-1) * integral from 0 to t of: 1/(z + 2) dz] + [integral from 0 to t of: 1/(z + 1) dz]
= limit as t-> infinity of: [(-ln|z + 2|) (evaluated at t and 0) + (ln|z + 1|) (evaluated at t and 0)]
= limit as t-> infinity of: [-ln|t + 2| + ln(2)] + [ln|t + 1| - 0]
If I solve the limit now it looks like I get: [-infinity + ln(2)] + [infinity]
But wouldn't this be divergent and equal to infinity? The book says it is convergent and equal to ln(2)...

 

[pka]

\(\displaystyle - \ln (t + 2) + \ln (2) + \ln (t) - \ln (1) = \ln \left( {\frac{t}{{t + 2}}} \right) + \ln (2)\)
Now reconsider the limit.

 

[galactus]

You can use algebra to expand your quadratic instead of going thorugh the A and B partial fraction thing.

\(\displaystyle \frac{1}{z^{2}+3z+2} = -\left[\frac{(z+1)-(z+2)}{(z+1)(z+2)}\right]=\frac{1}{z+1}-\frac{1}{z+2}\)

Next, \(\displaystyle \int_{0}^{L}\left[\frac{1}{z+1}-\frac{1}{z+2}\right]dz\)

\(\displaystyle =ln(2)+ln(L+1)-ln(L+2)\)

Now, \(\displaystyle \lim_{L\to{\infty}}\left[ln(2)+ln(L+1)-ln(L+2)\right]\)

\(\displaystyle \lim_{L\to{\infty}}\left[ln(2)+\underbrace{ln(\frac{L+1}{L+2})}_{\text{limit is 0}}\right]\)

 

[MarkSA]

Hmm so in this problem it's just a matter of simplifying in order to evaluate the limit the right way?

One question though, how would you tell that the lim as L-> infinity of: ln[(L + 1)/(L + 2)] goes to zero? It looks like it wouldn't be possible to determine by looking, since the top is infinity + 1 and the bottom infinity + 2? And then the ln throws another kink in it...

 

[galactus]

I'm sorry, I thought you could see it. Sorry to be presumptuous.

Well, with a little bit of this and that we can tell.

\(\displaystyle \underbrace{\lim_{L\to{\infty}}ln(\frac{L+1}{L+2})=ln(\lim_{L\to{\infty}}\frac{L+1}{L+2})}_{\text{yes, you can do that\\limit property}}=ln(\lim_{L\to{\infty}}\frac{1+\frac{1}{L}}{1+\frac{2}{L}})\)

Now, can you see it?. What does ln(1) equal?.

 

[MarkSA]

thanks, I see it now. I'd forgotten you could pull the ln outside of the limit.

 

[Unco]

thanks, I see it now. I'd forgotten you could pull the ln outside of the limit.

In general, if \(\displaystyle \displaystyle \lim_{x\to a} f(x)\) exists and \(\displaystyle g\) is continuous at the value of that limit, then \(\displaystyle \displaystyle \lim_{x\to a} g(f(x)) = g\left(\lim_{x\to a}f(x)\right)\).