Question

(dx/x^p) lower limit of integration 0 and upper limit of integration is 1. Find value of p which the integral converges and evaluate the integral for those value at p.

Answer 1

\[\int_0^1 \frac{dx}{x^p}\]
is this the question @mandonut

Answer 2

yes

Answer 3

so do you have an idea how to do this?

Answer 4

i got the integral which is 1/(1-p)(x^(p-1))

Answer 5

hmm

Answer 6

actually you need to turn this into
\[\large \int x^{-p} dx\]
then use power rule.

is that what you did?

Answer 7

yes

Answer 8

then your power rule should be \[\frac{1}{1 - p} \cdot x^{-p + 1}\]

Answer 9

or for organization purposes
\[\large \frac{x^{1-p}}{1-p}\]

Answer 10

which is the same as my answer if u factor -1 for the exponent of x

Answer 11

oh wait. x(p-1) is in the denominator?

Answer 12

then it's right

Answer 13

x^(p-1) *

Answer 14

so did you try plugging in the limits?

Answer 15

when 0 its undefined thats what im stuck on

Answer 16

actually when it's 1 then it's undefined

Answer 17

u plug 1 into x, not p

Answer 18

oops lol yeah

Answer 19

got confused

Answer 20

wait...how did it become undefined?

Answer 21

1/0^(p-1) and p cannot be equal to 1

Answer 22

that is why you should write it as \[\frac{x^{1-p}}{1-p}\]

Answer 23

see if x = 0 then it's just 0...as lng as p is not 1

Answer 24

good point

Answer 25

thanks man

Answer 26

you're welcome

Answer 27

It is undefined for that integral, but not because\[\frac{x^{1-1}}{1-1}\text{ is undefined, it is because}\]
\[\int_0^1\frac{dx}{x} = ln(x)\vert_0^1 = \ln(1)-\ln(0)\text{ and }\ln(0)\text{ is undefined}\]

Answer 28

\[lim_{\epsilon\rightarrow 0}\int_\epsilon^1\frac{dx}{x^p}\]
For p=1:\[lim_{\epsilon\rightarrow 0}\int_\epsilon^1\frac{dx}{x}=lim_{\epsilon\rightarrow 0}(\ln{x}\vert_{\epsilon}^1)=lim_{\epsilon\rightarrow 0}(\ln{1}-\ln{\epsilon})=0-lim_{\epsilon\rightarrow 0}(\ln{\epsilon})=\]\[0-(-\infty)=\infty\]For p<>1\[lim_{\epsilon\rightarrow 0}\int_\epsilon^1\frac{dx}{x^p}=lim_{\epsilon\rightarrow 0}\frac{x^{1-p}}{1-p}\vert_{\epsilon}^1\]\[=lim_{\epsilon\rightarrow 0}\frac{1}{1-p}-\frac{\epsilon^{1-p}}{1-p}=\frac{1}{1-p}-lim_{\epsilon\rightarrow 0}\frac{\epsilon^{1-p}}{1-p}=\frac{1}{1-p}(1-lim_{\epsilon\rightarrow 0}\epsilon^{1-p})\]
This limit only exists when 1-p is positive. Therefore p<1 and the integral =
\[\frac{1}{1-p}\]