Expert: Paul Klarreich - 6/25/2006

Question
Hi,

My name if Flori and I am trying to solve an improper integral, such as int(ln(x)/x^3, x = 1 .. infinity).

I did lim as b-->infinity from int(ln(x)/x^3, x = 1 .. b ).
I solved the integral by parts, I let my u=ln(x) and dv=x^3dx. My answer was :
    x^4/4 * (ln(x)-1/4) from 1 to b.

Now I need to take limit of this value and I am not sure how to do it. I got lim as b goes to infinity from : b^4/4*(ln(b)-1/4)-1/4(ln1-1/4).Now the part w/ b  I don't know where it goes. Is it a finite number or infiniy? The 2nd part is going to be 1/16.

Is my reasoning corect? How  do you take the limit of something that is of the form infinity*infinity-infinity?  

Answer
Hi, Flori,

Subject:  Calculus: convergent or divergent integrals
Question:  Hi,

My name if Flori and I am trying to solve an improper integral, such as int(ln(x)/x^3, x = 1 .. infinity).

I did lim as b-->inf from int(ln(x)/x^3, x = 1 .. b ).
I solved the integral by parts, I let my u=ln(x) and dv=x^3dx. My answer was :
   x^4/4 * (ln(x)-1/4) from 1 to b.

>> Oops --  your  'dv' is not  x^3 dx, it's  x^-3 dx.  That will mess things up.  See below.


Now I need to take limit of this value and I am not sure how to do it. I got lim as b goes to infinity from : b^4/4*(ln(b)-1/4)-1/4(ln1-1/4).Now the part w/ b  I don't know where it goes. Is it a finite number or infiniy? The 2nd part is going to be 1/16.

Is my reasoning corect? How  do you take the limit of something that is of the form infinity*infinity-infinity?  

>> That's a different question.  You will have all sorts of indeterminate forms, but you have to do some kind of algebraic transformation that produces:

infinity
--------
infinity

and then you can use l'Hospital's rule.  Sometimes the transformation is not obvious and you have to ask some very old person how to do it.

---------------------------
Integration by parts is a good idea, but it works like this:  WARNING: USE A FIXED SIZE FONT TO VIEW THIS.

{inf  ln x
|     ---- dx
}1     x^3

u = ln x,  because it is easy to differentiate  ln x
du = 1/x dx

dv = x^-3 dx,  because it is easy to integrate  x^-3
   x^-2   
v = -----
    -2

So your integral is:
 ln x     {  1    dx
-------  - | --- ----
- 2 x^2    }  x  -2x^2

 ln x     {   dx
-------  + | -------
- 2 x^2    }  2x^3


 ln x      x^-2  
-------  + ------
- 2 x^2      -4

1    ln x         1  
-- [ --------  + ------]  for  x = 1 to infinity.
-2    x^2         2x^2

At x = 1, this is -1/4  (ln 1 = 0, of course)

At x = infinity, both terms --> 0.  Use l'Hospital's rule for the ln x/x^2.

So the answer is +1/4