## sjg13e one year ago Improper Integral, (2 to infinity)

1. sjg13e

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

2. sjg13e

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

3. sjg13e

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

4. sjg13e

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?

5. SolomonZelman

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

6. SolomonZelman

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).

7. SolomonZelman

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

8. sjg13e

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

9. SolomonZelman

Yes

10. SolomonZelman

Comparison test, to a larger series.

11. SolomonZelman

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

12. SolomonZelman

If you want, we can solve the integral tho

13. sjg13e

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

14. SolomonZelman

$$\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v-9}{~\rm dv} }$$ $$\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v+16-16-9}{~\rm dv} }$$ $$\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+4)^2-25}{~\rm dv} }$$ set u=v+4 this is where I would start

15. freckles

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

16. freckles

you could use trig sub or do partial fractions

17. SolomonZelman

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

18. SolomonZelman

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

19. SolomonZelman

and do the partial fractions...

20. freckles

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

21. SolomonZelman

trig after u sub?

22. freckles

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

23. sjg13e

24. freckles

though I much prefer the partial fractions

25. SolomonZelman

Yeah, definitely... |dw:1443654159980:dw|

26. sjg13e

okay thank you! i messed up in the beginning with integration. i'll probably used partial fractions to solve it then

27. SolomonZelman

yes, just partial fractions for 1/((v-1)(v+9))

28. freckles

@sjg13e well you messed up before the integration you used that 1/(a+b+c)=1/a+1/b+1/c which is not true

29. sjg13e

yes, that's what i meant.

30. sjg13e

thank you all very much! i appreciate your help!

31. freckles

np :)

32. SolomonZelman

Σ = np × 2