convergence/divergence of int. of (ln x)/x^3 from 1 to infty

P

paulxzt

Junior Member
Joined
Aug 30, 2006
Messages
65
determine whether the integral is convergent or divergent. evaluate if it's convergent.

the integral of (ln x)/x^3 from 1 to infinity.

do I integrate by parts ?
thank you for any help
 
integral lnx /x^3 dx from 1 to oo

let u = lnx
then
du =1/x dx

let dv = x^-3 dx
then
v=-x^-2/2

integral = -[1/2]lnx [1/x^2] - integral [1,00 ][-1/[2x^2][dx/x]
integral= -[1/2x^2]lnx +[1/2] integral[1,00] [dx/x^3]
integral= -[1/2x^2]lnx +1/2[-1/[2x^2] evaluated from 1 to oo

at oo integral= 0-0
at 1 integral= 0-1/4

integral from 1 to infinity of lnx dx /x^3 = 1/4

please check for errors
Arthur
 
arthur ohlsten said:
integral lnx /x^3 dx from 1 to oo

let u = lnx
then
du =1/x dx

let dv = x^-3 dx
then
v=-x^-2/2

integral = -[1/2]lnx [1/x^2] - integral [1,00 ][-1/[2x^2][dx/x]
integral= -[1/2x^2]lnx +[1/2] integral[1,00] [dx/x^3]
integral= -[1/2x^2]lnx +1/2[-1/[2x^2] evaluated from 1 to oo

at oo integral= 0-0
at 1 integral= 0-1/4

integral from 1 to infinity of lnx dx /x^3 = 1/4

please check for errors
Arthur
Click to expand...

I got -1/4. how is it positive?
 
I did do it line by line.
I suggest you check your work to find where you missed a - sign.

I suggest you sketch the curve. At x=1 y=0, and at x=oo,y=0
take the derivative , set to 0 and find the point of maximum value
y=[x^-3]lnx
dy/dx= x^-3[1/x]+lnx [-3x^-4] set=0
0=1/x^4 -3lnx/x^4 cross multiply
0=1-3lnx
lnx=1/3
x=e^1/3

you will see the curve reaches a maximum,and then approaches 0 as x approaches infinity. The area is positive. Check your work for a error in sign.

by the way last night when I found the point of maximum value I got e. Please check my work, but in either case the area is positive.

Arthur
 
Arthur,

I was suggesting that the student show his/her work - line by line.
 
You must log in or register to reply here.
Top