# I dont get this convergent/divergent problem

1. Feb 8, 2006

### nick727kcin

please help. its 4 am and im still stuck on my math hw

instriuctions:
find the values of p for which the interval converges and evalaute the integral for those values of p

frankly, i dont really get this convergent/divergent fooey

here is the question

48. interval of 1/(x(lnx)^p) dx from e to infinity

ill draw it in a second

Last edited: Feb 8, 2006
2. Feb 8, 2006

### Tom McCurdy

lol nice sketch

use compartive p test to check answers

3. Feb 8, 2006

### Tom McCurdy

$$\int_{e}^{\infty} x dx = \frac{1}{xln(x)^p}$$

Will converge for p>1
Diverge for p less or equal 1

based on

$$\int_{1}^{\infty} x dx = \frac{1}{x^p}$$

4. Feb 8, 2006

### nick727kcin

thanks so much

5. Feb 8, 2006

### benorin

An improper integral is convergent if it is equal to a finite number, and divergent otherwise.

In your integral (by the way: nice sketch) put u = ln x so that du = 1/x dx and hence

$$\int_{x=e}^{\infty} \frac{1}{x(lnx)^{p}} dx = \int_{u=1}^{\infty} \frac{1}{u^{p}} du$$

Can you get it from there?

6. Feb 8, 2006

### nick727kcin

i got it now. thanks

:!!)

7. Feb 8, 2006

### nick727kcin

sorry to bother you guys, but im still a little confused

so all i do is find the integral of xlnx^-p?

(lnx)^p-1
---------
p-1

would that be all i have to do since i said that it has to be greater than 1?

and its still divergent by the way, even though i made p = 2

Last edited: Feb 8, 2006
8. Feb 8, 2006

### Tom McCurdy

$$\int_{e}^{\infty} \frac{1}{xln(x)^p}$$
Will converge to 1
when p=2

9. Feb 8, 2006

### Tom McCurdy

Like benorin said since you can use substitution
with u=ln(x) and du=dx/x
you will get the
Substitute in the bounds with u
Lower: u(e)=1
Higher: u(infinity)=infinity
So from
$$\int_{e}^{\infty} \frac{1}{xln(x)^p}$$

you get
$$\int_{1}^{\infty} \frac{1}{x^p}du$$
which is a definition of a ptest which defines that
$$\int_{e}^{\infty} \frac{1}{x)^p}dx$$
will Converge for p>1 and Diverge for p less than or equal to 1

Btw another way you can test when p=2 is the
$$\int_{1}^{\infty} \frac{1}{x^2}$$

$$\lim_{B->\infty}[ \frac{-1}{x} ]_{1}^{B}$$

=
-1/B- (-1/1)
the (-1/B) goes to zero so you have
0-(-1/1)=1

Last edited: Feb 8, 2006
10. Feb 8, 2006

### nick727kcin

thanks sooooo much. you guys just made my day

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