Question

Help with improper integral?

Answer 1

\[\int\limits_{2}^{\infty} (4)/(y^2+2y-3)\]

Answer 2

Factorize denominator first?

Answer 3

right away i know you can use lim as n approaches infinity and use PFD but once i get passed that and evaluate the limits get confusing

Answer 4

Right that'd be \[\int\limits_{2}^{\infty} (4)/(y+3)(y-1)\] and then use PFD for A = -1 and B = 1

Answer 5

then lim as n approaches infinity of \[\int\limits_{2}^{n} (1)/(y-1) - (1)/(y+3)\]

Answer 6

So then you can split both into their own integrals to get lim as n approaches infinity of [ln (n - 1) - ln(2 - 1)] - [ln (n+3) - ln (2+3)]

Answer 7

which i got lim as n approaches infinity of ln(n-1) - ln(n+3) + ln (5)

Answer 8

but thats as far as i got

Answer 9

@mathmale @phi some help, please?

Answer 10

@Mehek14 @jhonyy9 any ideas?? :o

Answer 11

ln(n-1) - ln(n+3) is equivalent to \[\ln \frac{ n-1 }{ n+3 }\]. What is the limit of that as n approaches infinity?

Answer 12

So, in the end, does your integral converge or diverge? If it converges, then to what value?

Answer 13

OH! Thank you. it was that property if logs i was missing. i forgot you can divide when their being subtracted. the ln(5) can just be moved out? So thats infinifty over infinity so we'd use L'hospitals rule then?

Answer 14

which would just be ln (1) which is 0? so ln(5) + 0?

Answer 15

I prefer to think big number / big number --> 1

Answer 16

ok "infinity"/"infinity". LOL

Answer 17

So then that'd mean it converges to ln(5)?

Answer 18

Yes, if y ou want to do the problem formally. In this case I'd just take a look at "n to the first power" in both numerator and den. and realize that this would approach 1 in the limit.

I don't care for that inf / inf; it's far too indefinite.

Why don't you follow your original impulse and apply l'Hop's rule?

You'll get 1 as the limit. As you already know, the ln of 1 is 0, leaving you with ln 5.

Answer 19

Thanks. I mean infinity over infinity definitley isnt "1" I just meant that because its in that form of infinite over infinite that we'd apply L'Hops which of course would just be ln(1) which is the same as zero and so ln(5) + 0 is just ln(5)

Answer 20

my professor bashed it through our heads that "infinity"/"infinity" is never 1 which is ofc true since infinity is just a concept

Answer 21

limit ln (stuff) = ln ( lim stuff)

in this case (n+1)/(n+3) = 1 -4/(n+3)

\[ \frac{n+1}{n+3} = 1 - \frac{4}{n+3} \]
and the limit is evidently 1 as n-> infinity

Answer 22

right. thats another rule of limits/logs i forgot. thank you! so we can just pull the limit into the stuff of the limit and just apply l'h or divide by the highest power of n?

Answer 23

into the stuff of the ln*

Answer 24

thanks again all btw! almost done with calc 2. almost...

Answer 25

Happy to be of help! Come back with more questions.

Answer 26

Most definitely thank you sir