• Dec 17th 2012, 04:28 PM
skinsdomination09
integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATION)
I integrated the function to become

ln x / (x^2)

u = lnx
du = 1/x
dv = 1/x^2
v = -1/x

UV - int VDU

lnx * -1/x - - int of 1/x^2 becomes -1/x

so

- lnx/x - lnx (or close to this)

• Dec 17th 2012, 04:58 PM
Plato
Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATIO
Quote:

Originally Posted by skinsdomination09
I integrated the function to become

ln x / (x^2)

u = lnx
du = 1/x
dv = 1/x^2
v = -1/x

UV - int VDU

lnx * -1/x - - int of 1/x^2 becomes -1/x

so

- lnx/x - lnx (or close to this)

Look at this webpage.
• Dec 17th 2012, 05:08 PM
skeeter
$\int_1^\infty \frac{\ln{x}}{x^2} \, dx$

$u = \ln{x}$ ... $du = \frac{1}{x} \, dx$

$dv = \frac{1}{x^2} dx$ ... $v = -\frac{1}{x}$

$\lim_{b \to \infty} \int_1^b \frac{\ln{x}}{x^2} \, dx = \lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \int_1^b -\frac{1}{x^2} \, dx$

$\lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \left[\frac{1}{x} \right]_1^b$

$\lim_{b \to \infty} \left[-\frac{\ln{b}}{b}\right] - \left[\frac{1}{b} - 1 \right]_1^b = 1$
• Dec 17th 2012, 05:09 PM
skinsdomination09
Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATIO
Quote:

Originally Posted by Plato

delete
• Dec 17th 2012, 05:13 PM
skinsdomination09
Quote:

Originally Posted by skeeter
$\int_1^\infty \frac{\ln{x}}{x^2} \, dx$

$u = \ln{x}$ ... $du = \frac{1}{x} \, dx$

$dv = \frac{1}{x^2} dx$ ... $v = -\frac{1}{x}$

$\lim_{b \to \infty} \int_1^b \frac{\ln{x}}{x^2} \, dx = \lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \int_1^b -\frac{1}{x^2} \, dx$

$\lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \left[\frac{1}{x} \right]_1^b$

$\lim_{b \to \infty} \left[-\frac{\ln{b}}{b}\right] - \left[\frac{1}{b} - 1 \right]_1^b = 1$

Makes more sense, Thanks But wouldn't the ln b / b (since it's approaching infinity) be indetirminate so it would be divergant. My original mistake was that I integrated the second half twice by accident, but I still don't get how you can just "ignore" ln b /b since ln(infinity) doesn't equal 0 or anything.

Oh... is it bc the top is growing slower than the bottom so its just a really small number over a big one and thus goes to zero? Simple Alg. + early Calc skills holding me back(Angry)
• Dec 17th 2012, 05:17 PM
skeeter
L'Hopital ...

$\displaystyle \lim_{b \to \infty} \frac{\ln{b}}{b} = \lim_{b \to \infty} \frac{1}{b} = 0$
• Dec 17th 2012, 05:19 PM
skinsdomination09
$\displaystyle \lim_{b \to \infty} \frac{\ln{b}}{b} = \lim_{b \to \infty} \frac{1}{b} = 0$