Improper Integral Convergence for f(x)=1/(x^p)

[zachem62]

Homework Statement

Consider the function f(x)=1/(x^p).

When p>1, the integral of 1/(x^p) from 1 to infinity converges.


i) For what values of p does the integral of 1/(x^p) from 0 to 1 converge? (0<p<infinity, p does not equal 1).

ii) Confirm the answer by re-writing the integral of 1/(x^p) from 0 to 1 in terms of area and an integral in terms of y. Comment on any symmetry/asymmetry that this relation demonstrates).

I did a question similar to this but it was much simpler. I have no idea how to even start with this question. PLEASE HELP!

The Attempt at a Solution

 

[LCKurtz]

Just calculate $\int_\epsilon^1 x^{-p}\, dx$, let $\epsilon \rightarrow 0^+$ and see which $p$ values work.

 

[iknowless]

i)(0<p<infinity, p does not equal 1).

Is this your answer to part one? If so, consider that these values will make the integral diverge, not converge. Also, please show us your work.

\int_{0}^{1} x^{-p} dx = ?

 

[zachem62]

Is this your answer to part one? If so, consider that these values will make the integral diverge, not converge. Also, please show us your work.

\int_{0}^{1} x^{-p} dx = ?

No that is not the answer to part one. Everything I have posted is part of the question and I have no clue how to get started and finish the question.

 

[LCKurtz]

No that is not the answer to part one. Everything I have posted is part of the question and I have no clue how to get started and finish the question.

Did you read post #2? That gives a clue.

 

[iknowless]

On the interval [0,1], which is where we are evaluating the integral, are there any points of discontinuity for values of x?

If so, then the integral is improper, and you must replace that x-value with a variable, then take the limit as that variable approaches the x-value from either the right or the left. Your book will have a specific theorem or example of an improper integral.