Question

integral from 0 to 1 for lnx/x dx

Answer 1

Please help!

Answer 2

Since \[\frac{lnx}{x}\] doesn't seem to have an elementary anti-derivative, we have to mess around with it. The first thing I always see is if there is a u-substitution that turns it into something I know. Any ideas what you could substitute?

Answer 3

yah so we would let u=lnx
du=1/x dx

Answer 4

then we have integral of 0 to 1 of u

Answer 5

Then what would that make your integral look like?

Answer 6

then we have u^2/2
but im lost from here
i know it is divergent but i dont know how to show that

Answer 7

We have \[\frac{u^2}{2} + c\] But it isn't from 0 to 1. Because\[u = lnx, x=0, u = -\infty, x=1, u = 0\]
If we use these limits, what does that say about the divergence?

Answer 8

well so, im lost sorry
so what im doing is [ln^2(1)/2 - ln^2(0)/2]
and then it is [0-und.] so it makes it undefined?

Answer 9

this is plugging u=lnx back into antideriv

Answer 10

If you have any discontinuities in the function, in this case \(\ln(x)/x\) is not continuous at \(x=0\), then you have to use a limit to evaluate it.

Answer 11

ok so i would use the lim as x aproachs 0 ?

Answer 12

A one sided limit, depending on which integration limit we are talking about.

Answer 13

Since 0 is a lower limit, we would do \(x\to 0^{-}\)

Answer 14

This is the part of calculus that I haven't touched in eight years, a bit rusty on it.

Answer 15

ok im going to do so
its ok as atleast you tried helping thats really nice!

Answer 16

\[
\lim_{x\to 0^-}\frac{\ln^2(1) -\ln^2(x)}{2}
\]

Answer 17

ok so would i do lim as x aproaches 0^-
integral of x to 1 of lnx/x

Answer 18

It simplifies to \[
\dfrac{0-\lim_{x\to 0^-}\ln^2(x)}{2}
\]

Answer 19

this graph doesnt exists from 0^- direction

Answer 20

http://www.wolframalpha.com/input/?i=lim_%7Bx+to+0%5E-%7D+ln%5E2%28x%29

Answer 21

hmm, actually maybe I should have said \(0^+\)

Answer 22

ok that i think is better cause the graph doesnt exists its domain is (0,infin)

Answer 23

In either case, this integral results in \(-\infty\)

Answer 24

isnt it positive infinity?

Answer 25

\[
\frac{0-\infty}{2} \to -\infty
\]

Answer 26

ok
thnks