# Improper Integral's & L'Hopital's Rule

• Feb 21st 2011, 08:36 PM
rawkstar
Improper Integral's & L'Hopital's Rule
Hi. Im having trouble with this improper integral
it is the integral of 5/(z^2+7z+6)dz from 5 to +infinity

I know that you let t=+inf and set it up as the limit of the integral as t approaches +inf and solve

I am stuck at the part after you integrate & plug +inf back in for t & I know that you're supposed to use L'hopitals rule somewhere but am having trouble doing this

• Feb 21st 2011, 08:39 PM
Random Variable
Did you use partial fractions?
• Feb 21st 2011, 11:54 PM
CaptainBlack
Quote:

Originally Posted by rawkstar
Hi. Im having trouble with this improper integral
it is the integral of 5/(z^2+7z+6)dz from 5 to +infinity

I know that you let t=+inf and set it up as the limit of the integral as t approaches +inf and solve

I am stuck at the part after you integrate & plug +inf back in for t & I know that you're supposed to use L'hopitals rule somewhere but am having trouble doing this

You do not need L'Hopital it is a perfectly ordinary limit problem. Post what you have and we will advise you on how to proceed.

CB
• Feb 22nd 2011, 02:22 AM
chisigma
If f(z) has only 'simple poles' $\alpha_{1}, \alpha_{2}, ..., \alpha_{n}$ , then is...

$\displaystyle f(z)= \sum_{n=1}^{n} \frac{r_{n}}{z-\alpha_{n}}$ (1)

... where...

$\displaystyle r_{n}= \lim_{z \rightarrow \alpha_{n}} f(z)\ (z-\alpha_{n})$ (2)

Now it is evident that the limits (2) are 'ideterminate forms' of the type $\frac{0}{0}$ so that they can be solved using l'Hopital's rule...

Kind regards

$\chi$ $\sigma$
• Feb 22nd 2011, 04:25 AM
CaptainBlack
Quote:

Originally Posted by chisigma
If f(z) has only 'simple poles' $\alpha_{1}, \alpha_{2}, ..., \alpha_{n}$ , then is...

$\displaystyle f(z)= \sum_{n=1}^{n} \frac{r_{n}}{z-\alpha_{n}}$ (1)

... where...

$\displaystyle r_{n}= \lim_{z \rightarrow \alpha_{n}} f(z)\ (z-\alpha_{n})$ (2)

Now it is evident that the limits (2) are 'ideterminate forms' of the type $\frac{0}{0}$ so that they can be solved using l'Hopital's rule...

Kind regards

$\chi$ $\sigma$

Now you are making things more complicated than they need be. You end up with the log of a rational function (from your equation (1)) and that can be handled perfectly well without resort to M. L'Hopital.

CB