Question

Improper Integral, (2 to infinity)

Answer 1

\[\int\limits_{2}^{\infty}\frac{ 1 }{ v^2 + 8v - 9 }dv\]
Does it converge or diverge?

Answer 2

\[\int\limits_{}^{}\frac{ 1 }{ v^2 + 8v - 9 }dv = \frac{ -1 }{ v } + \frac{ 1 }{ 8 }\ln|v| - \frac{ 1 }{ 9 }v\] (this is what I got)

Answer 3

\[\lim_{t \rightarrow \infty}[\frac{ -1 }{ v }+\frac{ 1 }{ 8 }\ln|v|-\frac{ 1 }{ 9 }v]\]
where a = 2 and b = t

Answer 4

I got \[\frac{ 1 }{ 3 }-\frac{ 1 }{ 8 }\ln|2|\] as my answer. But it's wrong.
I know it converges, maybe I messed up with inputting the values?

Answer 5

Suppose you want to know whether
\(\large\color{black}{\displaystyle\sum_{n=7}^{\infty}\frac{1}{n^2} }\)
converges or not.

you know that converges, (to π²/6 by Euler)

And a bigger series
\(\large\color{black}{\displaystyle\sum_{n=7}^{\infty}\frac{1}{n^2+8v-9} }\)
would all the more so converge

Answer 6

So if the first series converges (and must be - so does its integral)
THEN, all the more so the second series (and must be - so does the integral).

Answer 7

That is ian observation if you studied about the *integral test*... have you?

Answer 8

Okay, yes. That is also expressed by the comparison test, right?

Answer 9

Yes

Answer 10

Comparison test, to a larger series.

Answer 11

It should really start from n=2, not n=7, but that doesn't matter.

Answer 12

If you want, we can solve the integral tho

Answer 13

okay thank you, id be awesome if you helped me solve it

Answer 14

\(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v-9}{~\rm dv} }\)

\(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v+16-16-9}{~\rm dv} }\)

\(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+4)^2-25}{~\rm dv} }\)

set u=v+4
this is where I would start

Answer 15

I think I see what you did...
you think \[\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \text{ but this is not true }\]

Answer 16

you could use trig sub or do partial fractions

Answer 17

yes, partical fractions from the beginning is better I guess, because with my way you still have partial fractions

Answer 18

\(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+9)(v-1)}{~\rm dv} }\)

Answer 19

and do the partial fractions...

Answer 20

@SolomonZelman I think you could have done trig sub without partial fractions
I was just offering another route

Answer 21

trig after u sub?

Answer 22

\[\int\limits \frac{1}{(v+4)^2-25} dv \\ \int\limits \frac{1}{25[\frac{1}{25}(v+4)^2-1]} dv \\ \frac{1}{25} \int\limits \frac{1}{(\frac{v+4}{5})^2-1} dv \\ \text{ recall } \tan^2(x)=\sec^2(x)-1 \\ \text{ so use sub } \frac{v+4}{5}=\sec(x) \\ \frac{1}{5} dv=\sec(x) \tan(x) dx \\ \frac{1}{25} \int\limits \frac{1}{\sec^2(x)-1} 5 \sec(x) \tan(x) dx \\ \frac{1}{25} \int\limits \frac{5 \sec(x)\tan(x)}{\tan^2(x)}dx\]
unless... I did something wrong we don't need partial fractions for the trig route here

Answer 23

Sorry, my wifi was spotty. I'll read through your explanation now

Answer 24

though I much prefer the partial fractions

Answer 25

Yeah, definitely...