Question

Calculus improper integral....

Answer 1

\[\int\limits_{1}^{\infty} \frac{ 1 }{ x ^{p}} dx \] where x is a positive real number
how do i find if it converges and its value or diverges

Answer 2

@Kainui

Answer 3

Do you mean p is a positive real number?

Answer 4

yes

Answer 5

whats stopping you from evaluating the integral ?

Answer 6

Well, we can evaluate it exactly if you watch out for when p=1

Answer 7

i know it converges if p>1 and diverges if p<=1 but i don't know why

Answer 8

what would the first step be?

Answer 9

maybe write it like this:
\[\int_0^\infty x^{-p}dx\]

Answer 10

if the limit when x approaches infinity equals 0, then it converges right?

Answer 11

Treat p like a constant

Answer 12

So do the power rule in reverse to integrate

Answer 13

add 1 to p then divide ...

Answer 14

\[\int\limits x ^{-p}dx \rightarrow \frac{ x ^{1-p} }{ 1-p}+c\]

Answer 15

Just evaluate the integral.

Answer 16

Given it's limits

Answer 17

\[[\frac{ \infty^{1-p} }{ 1-p }] - [\frac{ 1^{1-p} }{ 1-p }]\]

Answer 18

What is that equivalent to for any positive real value p?

Answer 19

\[\infty - \frac{ 1 }{ (1-p)^{1-p} }\]

i dont know :(

Answer 20

Well at p=1 you have you're dealing with undefined terms can't divide by zero
if p is less then 1 more than zero or more than 1

The first term is some function of infinity
second term is a real number

Answer 21

some function of infinity minus a real number is the solution

Answer 22

What do you know about convergent series and the rules for them regarding integrals?

Answer 23

no

Answer 24

What do we know?

Answer 25

i havent learned that

Answer 26

what do you mean by convergent series?

Answer 27

Your integral can be rewritten:

Answer 28

\[\zeta(x)=\sum_{n=1}^{\infty}(1/n^x)\]

Answer 29

This the Reimann-zeta function. Zeta of x is the the sum of 1 over n to power of x.

Answer 30

ok

Answer 31

im sorry, im not sure what to do

Answer 32

Have you learned about the p-series?

Answer 33

nope

Answer 34

Ok, I found it.. Your integral is equal to\[\sum_{n=1}^{\infty}(\frac{ 1 }{ n })^{c}\]

Answer 35

That can recognized as a geometric series
whose base is less than 1 absolutely therefore the sum is convergent and thus the integral is also convergent

Answer 36

oh ok. so thats a rule?

Answer 37

That's a property of geometric series.

Answer 38

And the second part is a property of the integral test.

Answer 39

Also you may recognize that the limit of the function involved is zero as x approaches infinity, which is another thing unique to convergent series.

Answer 40

how is the limit zero?

Answer 41

\[\lim_{n \rightarrow \infty}\left( \frac{ 1 }{ x } \right)^{n}\]

and
\[\lim_{x \rightarrow \infty } (\lim_{n \rightarrow \infty}(1/n)^{x})=0\]

Answer 42

ah ok. so now i evaluate that integral?

Answer 43

i still dont know how to evaluate \[\frac{ x ^{1-p} }{ 1-p}\]from 1 to infinity

Answer 44

what is going from 1 to infinity?

Answer 45

\[\int\limits_{1}^{\infty} \frac{ 1 }{ x ^{p} }dx = \frac{ x ^{1-p} }{ 1-p } [1,\infty]\]

Answer 46

it depends on your p value..

Answer 47

p>1 then it converges
p< or = 1 then it diverges (p series properties)

Answer 48

so if p>1, how can i evaluate it?

Answer 49

By first rewriting it as a geometric series

Answer 50

\[\sum_{n=1}^{\infty}ar ^{n-1}=(\frac{ a }{ 1-r }) \] when 0<r<1

Answer 51

You say that p is more than 1 but is p a constant or a variable?

Answer 52

p is a constant

Answer 53

Call n-1 your p

r will always be less than 1 in absolute value right?

Answer 54

yes