Question

\[\int\limits_{1}^{\infty} \left( \frac{\ln(x)}{x} \right)^{2011}dx\]

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

hey is this like a, whats it called, published problems or something like that?
like i see 2011
i done a 2007 problem in 2007 lol

Answer 3

oh haha its from a math tournament in 2011

Answer 4

probly why they chose that power

Answer 5

i should show you the problem i did in my senior seminar class

Answer 6

id like to see it if you have it

Answer 7

\[\lim_{n \rightarrow \infty}(1^\frac{1}{n}+2^\frac{1}{n}+...+2006^\frac{1}{n}-2007)\]

Answer 8

no power on 2007 right?

Answer 9

oops to the nth power

Answer 10

the whole thing is to the n powee
2007 is the 1st power

Answer 11

\[\lim_{n \rightarrow \infty}(1^\frac{1}{n}+2^\frac{1}{n}+...+2006^\frac{1}{n}-2007)^n\]

Answer 12

well i think its going to take some thought

Answer 13

im trying some e^ln(...) things

Answer 14

i think yours is going to take some thought as well

Answer 15

woops just broke maple. I tried to define the function in the limit using a for loop

Answer 16

the integral i posted doesnt work in mathematics software either

Answer 17

how long did it take you to solve this problem?

Answer 18

a semester lol

Answer 19

i didn't really solve it. i had help from the instructor but i understand it now

Answer 20

yikes well then let me know how to start it haha

Answer 21

you will need to use or i used
Let T(x)=a^x
T'(0)=lim x->0 (f(x)-f(0))/(x-0)=(a^x-1)/x=lna.
this should really help you get started i think.

Answer 22

let x=1/n
then n goes to infinity

Answer 23

i forgot to type x->0 on that one part if you didn't notice
lim x->0 (a^x-1)/x=lna

Answer 24

have you actually done the problem you typed above?

Answer 25

yeah i have

Answer 26

how long did it take you

Answer 27

about an hour of doing wrong things and then I did something that worked.

Answer 28

lol

Answer 29

do you want the general method to use?

Answer 30

let me look at it for a few seconds
i know how to integrate lnx/x using substition
something like this looks like we have to use integration by parts
it will be long but i could probably write it as a summation after figuring out the pattern
and then figure out the limit of the summation

Answer 31

i have wrote it as int(u^2011*e^(-2010u) du)

Answer 32

i did that too I couldn't get much further that way though

Answer 33

there is a better way?

Answer 34

yes substitution doesn't work, integration by parts works out very well

Answer 35

skip the substitution and go straight for the integration by parts?

Answer 36

instead of using both?

Answer 37

yeah just int by parts, and if you pick the correct u, dv there's a really nice pattern so you dont have to do it 2011 times

Answer 38

lol yes 2011 is basically infinity (countable)

Answer 39

haha yeah it would be no fun, aka maple gave up after 5 minutes

Answer 40

let maple run all night
and it didn't ever give me a limit for that 2007 problem

Answer 41

haha computers arent that smart yet

Answer 42

theyre pretty smart though

Answer 43

ok well I should get to bed, talk to you later

Answer 44

me too someone is waiting on me lol

Answer 45

good night :)

Answer 46

gn

Answer 47

i got -2011!*e/(2010^2010)

Answer 48

i think this is right

Answer 49

oops its divergent
sorry

Answer 50

i wrote it from 0 to infinity instead
of 1 to infinity
so we just have to look at lnb as b goes to infinity lnb->infinity

Answer 51

\[u=\ln(x)^n \rightarrow du=\frac{n*\ln(x)^{n-1}}{x}\]
\[dv=\frac{1}{x^n} \rightarrow v=-\frac{1}{(n-1)x^{n-1}}\]
so our integral is:
\[\frac{\ln(x)^n}{(n-1)x^{n-1}}+\frac{n}{n-1}\int\limits_{1}^{\infty}\frac{\ln(x)^{n-1}}{x^n}\]

The first term goes to zero.

the second interal we can perform the same integration by parts until we get:

\[\frac{n!}{(n-1)^n}\int\limits_{1}^{\infty}\frac{1}{x^n}\]
this integral is 1/(n-1) so finally:

\[\int\limits_{1}^{\infty}\left( \frac{\ln(x)}{x} \right)^n=\frac{n!}{(n-1)^{n+1}}=\frac{2011!}{2010^{2012}}\]

the general version is true when n>0

Answer 52

sorry, when n>1