Question

integration check

Answer 1

\[\int\limits_{0}^{\infty}\frac{dx}{\sqrt{lnx}}\]
what i did:
let u=-lnx
x=e^-u
dx=-e^-u du
when x=0 --> y=infinity
when x=1 --> y=0
\[\int\limits_{0}^{\infty}y^{-\frac{1}{2}}e^{-y}dy=\gamma (\frac{1}{2})=\sqrt{\pi}\]

Answer 2

its u instead of y :S

Answer 3

let u=-lnx ???

Answer 4

yeah

Answer 5

that would be square root of negative ...

Answer 6

why squareroot? i dont need a sqrt?

Answer 7

\( \frac{1}{\sqrt{\ln x}} = \frac{1}{\sqrt{-u}} \) ??

Answer 8

yeah but i didnt do that, i used another way

Answer 9

thts why x=e^-u

Answer 10

\[\ln(x)<0\text{ for }0<x<1\]

therefore \(\sqrt{\ln(x)}\) is not a real number.

Answer 11

did you want this?

\[\int\limits_{0}^{1}\frac{1}{\sqrt{-\ln(x)}}dx=\sqrt{\pi}\]