Question

(Integral from 1 to infinity) [ln(x)]/[x^2] Convergent or Divergent

Answer 1

\[\int\limits_{0}^{\infty} \frac{ \ln(x) }{ x ^{2} }dx\]

Answer 2

i let u = lnx
du = 1/x dx

and dv = \[x^{-2}\]
and then v = -x right??????

Answer 3

First, you can compute the integral by using integration by parts. If you use int udv = uv - int vdu and choose ln x as u, you should be fine.

Answer 4

im stuck on v, after that i can compute the improper integral

Answer 5

dv = x^(-2) dx (don't forget the dx). Also, you're having a little trouble with v. First, get the right integral for v, you're not right on that yet.

Answer 6

With getting integrals on the form x^n, you increase n by 1. That hint should get you going on the right track.

Answer 7

so -x^-1???

Answer 8

Yes, you're getting it now. Good job. Continue on with the parts aspect and ask a question if you get stuck.

Answer 9

ok thankyou

Answer 10

Once you get the indefinite integral, you can use L'Hopital's rule for limits. That will give you your answer.

Answer 11

\[-x ^{-1} - \int\limits_{1}^{\infty} -x ^{-1}(\frac{ 1 }{ x })dx\]

Answer 12

correct so far?

Answer 13

i mean -x^-1(lnx) then integral

Answer 14

Yes. I was writing a response pointing out the error that you caught yourself. You're still on the right track.

Answer 15

do you have to do another integration by parts for the -x^-1(1/x) inside the integral?

Answer 16

do you mind just showing me the steps? it takes a while for me to use the equation thing on here and type all of it out

Answer 17

You don't have to do a second integration by parts, there you are just simplifying exponential powers of x. That last integral will be very easy. Just do exponent arithmetic and you will be doing integration on x to a certain power. Tell me if that is a big enough hint or if you still need the steps on that.

Answer 18

it just simplifies to -1/x^2 right

Answer 19

Yes, and then just do the integral. Remember, the second half of the formula is - int v du. So, with two negative signs, you have, as the second term, adding the int x^(-2) dx. So, watch the +- sign

Answer 20

are you looking to just check for convergence/divergence as your original post indicates?

Answer 21

yes but i have to show why its convergent/divergent

Answer 22

\[\ln(x)<\sqrt{x}\]

Answer 23

use a comparison test

Answer 24

so it is converent then

Answer 25

yes

Answer 26

cool, i still have to evaluate the integral

Answer 27

a very simple integral

Answer 28

No, you can't just do the comparison test. You have to get the indefinite integral and apply the limits on the answer to the integral.

Answer 29

thats given me more problems than any other problem

Answer 30

im just rusty, were doing laplace transform and havent done improper integrals in a while

Answer 31

\[0\le\int\limits_{1}^{\infty} \frac{ \ln(x) }{ x ^{2} }dx\le \int\limits_{1}^{\infty} \frac{ \sqrt{x} }{ x ^{2} }dx\]

\[=\int\limits_{1}^{\infty} x^{-3/2}dx=\cdots\]

Answer 32

I = -x^(-1)(lnx + 1). Using L'Hopital and getting the limit of the derivative of the numerator over the denominator will show that lim as x-> inf of (ln x)/x goes to zero.

Answer 33

Eventually, you will get as the integral, -x^(-1)(lnx + 1). I know you don't need the exact value for the integral, but the evaluation will give you an actual finite value which will then answer your question on convergence/divergence.

Answer 34

im sorry i just really am confused now...

before zarkon came in i got to \[-\infty ^{-1}(\ln \infty) - (-0^{-1}(\ln 0) + -\frac{ 1 }{ 3 }x ^{-3}\]

Answer 35

and im not confident in that at all

Answer 36

i have to evaluate the -1/3x^-3 from infinity - 0 too

Answer 37

Don't worry, I'll get you through the rest of it.

Answer 38

thankyou

Answer 39

First, did you get as the final integral -x^(-1)(lnx + 1)?

Answer 40

no i just got -x^-1(lnx)

Answer 41

i never had a +1 in mine

Answer 42

are you going from 1 to infinity or

0 to infinity?

Answer 43

1 to infinity.. i accidentally put 0 in the opening equation

Answer 44

The 1 is from doing - int v du. I'll show you the steps. You've worked way hard on this and you have done well, so I'm not really giving away the store here. You did a tremendous amount of work.

Answer 45

ok...carry on ;)

Answer 46

I = -x^(-1)lnx + int x^(-2) dx. the integration by parts is uv - int v du, but you have a negative in v, so there's a double negative going on there that I just combined.

Answer 47

i never set it to equal 1 like that

Answer 48

i have that on my paper

Answer 49

That -x^(-1)lnx (first term) is uv. So, I = -x^(-1)lnx - x^(-1) or -x^(-1)(ln x + 1). Limit of that as x -> inf is the limit of the derivative of the numerator over the derivative of the denominator (L'Hopital). So, limit of (-1/x)/1 as x -> inf, so that would be 0. So, all you have to do is evaluate at 1 (subtracting, again a double minus). So, x^(-1)(ln x + 1) (notice no negative now as we subtracted a negative) at 1 is 1^(-1)[(ln 1) + 1] = 1. Convergent.

Answer 50

Sweet thank you for all your help, I really appreciate that

Answer 51

I hope I was helping. It's hard to show so many steps and not lose someone. I really do hope this has lifted a burden for you and that you get what I was trying to say. If not, I blame myself as teacher. You obviously are a good and hard-working student. Good luck to you.

Answer 52

I completely get what you were trying to say. I got lost there for a second but once you typed out the final steps it made sense. thanks again

Answer 53

You have a wonderful day. and you be a good engineer or whatever!