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)

So where the hell does this mystical one come from? It looks as if it'd be divergant from this!!! Please help!!!!!!! (Headbang)(Headbang)(Headbang)

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)

So where the hell does this mystical one come from? It looks as if it'd be divergant from this!!! Please help!!!!!!! (Headbang)(Headbang)(Headbang)

Look at this webpage.

Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

$\displaystyle \int_1^\infty \frac{\ln{x}}{x^2} \, dx$

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

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

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

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

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

Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATIO

Quote:

Originally Posted by

**Plato**

delete

Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

Quote:

Originally Posted by

**skeeter** $\displaystyle \int_1^\infty \frac{\ln{x}}{x^2} \, dx$

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

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

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

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

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

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)

Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

L'Hopital ...

$\displaystyle \displaystyle \lim_{b \to \infty} \frac{\ln{b}}{b} = \lim_{b \to \infty} \frac{1}{b} = 0$

Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

Quote:

Originally Posted by

**skeeter** L'Hopital ...

$\displaystyle \displaystyle \lim_{b \to \infty} \frac{\ln{b}}{b} = \lim_{b \to \infty} \frac{1}{b} = 0$

Wow. I'm an idiot lol. Thanks