Question

Problem from someone here (I forgot who)
Compute
\[\int_1^\infty \left(\frac{\ln x}{x}\right)^{2001}\]

Answer 1

Let's make it 2011

\[\int_1^\infty \left(\frac{\ln x}{x}\right)^{2011}\]

Answer 2

put lnx=y

Answer 3

if y = ln(x) then e^y = x.
change limits when x = 1, y = 0 and when x = infty, y = infty

Answer 4

so
\[\int_0^\infty y^ke^{-ky} dy\]

Answer 5

how can it be infinity....lnx/x always <1

Answer 6

looks like a gamma distribution

Answer 7

Good cruffo :)

Answer 8

im a noob. wow like a week out from BC Calc and I forget everything :(

Answer 9

woho!

Answer 10

knowing gamma function will help alot but if you don't know it you can still solve the problem though.

Answer 11

yah, by parts - just need to figure out the pattern.

Answer 12

http://cache.ohinternet.com/images/f/f3/Fanserviceftw_7538_Obvious_troll_is_obvious.jpg = Lovelesstime

Answer 13

By parts and looking for a pattern in the outcomes,
\[\frac{(k-1)!}{k^k}\]

Answer 14

yes..gamma functions are like that...

Answer 15

this is not actually gamma type because e^-ky is here..in gamma the k=1

Answer 16

I think I remember do all sorts of fancy footwork to change the variables so you could use the distribution and call it a day. Been a few years - have to check my notes!