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.