1. ## Simple integral

Simple integral from 1 to infinity of (ln x)/(x^2).

2. Originally Posted by Possible actuary
Simple integral from 1 to infinity of (ln x)/(x^2).
We proceed By Parts:

Again, this is an improper integral

lim{N-->infinity} int{lnx/x^2}dx from 1 to N = -x^-1*lnx + int{x^-1 * x^-1}dx
.....................= lim{N--> infinity} -lnx/x - x^-1 + C
.....................= lim{N-->infinity} -lnN/N - 1/N + ln1 + 1
.....................= 0 + ln1 + 1
.....................= ln1 + 1
.....................= 1.................................since ln(1)=0 (thanks ecMathGeek)

So i went through this kind of fast neglecting the finer details. ask away if you're unclear about something

You know how to do integration by parts right?

3. Originally Posted by Possible actuary
Simple integral from 1 to infinity of (ln x)/(x^2).
Use Integration by Parts on this:
INT (udv) = uv - INT(vdu) ... I'm hoping you're familiar with this.

At the end, we will also need to use limits in order to integrate from 1 to infinity.

Let u = lnx <--> du = 1/xdx
Let dv = 1/x^2dx <--> v = -1/x

So, INT lnx/x^2 dx becomes:
-lnx/x - INT (-1/x^2)
-lnx/x - 1/x
-(lnx + 1)/x

Now, plugging in the limits of integration, using a limit for the upper limit of infinity:

lim[n->infinity] -(lnx + 1)/x from (1 to n)
lim[n->infinity] -(ln(n) + 1)/n + (ln(1) + 1)/1

Now we need to use L'Hopitals to simplify the limit

lim[n->infinity] (-1/x)/1 + 1 = 0/1 + 1 = 1

4. Originally Posted by ecMathGeek
Use Integration by Parts on this:
INT (udv) = uv - INT(vdu) ... I'm hoping you're familiar with this.

At the end, we will also need to use limits in order to integrate from 1 to infinity.

Let u = lnx <--> du = 1/xdx
Let dv = 1/x^2dx <--> v = -1/x

So, INT lnx/x^2 dx becomes:
-lnx/x - INT (-1/x^2)
-lnx/x - 1/x
-(lnx + 1)/x

Now, plugging in the limits of integration, using a limit for the upper limit of infinity:

lim[n->infinity] -(lnx + 1)/x from (1 to n)
lim[n->infinity] -(ln(n) + 1)/n + (ln(1) + 1)/1

Now we need to use L'Hopitals to simplify the limit

lim[n->infinity] (-1/x)/1 + 1 = 0/1 + 1 = 1
i think you forgot the ln(1) at the end friend...or did i mess up again?

5. Originally Posted by Jhevon
i think you forgot the ln(1) at the end friend...or did i mess up again?
I was just checking over our work to see if either of us messed up. I'm not sure just yet, but in case you forgot, ln(1) = 0

6. Originally Posted by ecMathGeek
I was just checking over our work to see if either of us messed up. I'm not sure just yet, but in case you forgot, ln(1) = 0
yup, you're right. i messed up again silly me

7. Originally Posted by Jhevon
i think you forgot the ln(1) at the end friend...or did i mess up again?
You didn't state that you used L'Hopital's Rule on lim[n->infinity] of ln(n)/n. But besides that, our work is the same.

8. Originally Posted by ecMathGeek
You didn't state that you used L'Hopital's Rule on lim[n->infinity] of ln(n)/n. But besides that, our work is the same.
Yeah, i kind of did it intuitively. i know that N increases a lot faster than ln(N) so the fraction would go to 0. but you're right, i should have done L'Hopital's for the benifit of the class