Find the values of p for which the integral converges?

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SUMMARY

The integral ∫ 0->1 1/(x^p) dx converges for values of p less than 1. Evaluating the integral for p < 1 yields the result 1/(1-p). For p equal to or greater than 1, the integral diverges, as demonstrated by limits approaching zero. The discussion highlights the importance of understanding the behavior of the function f(x) = 1/x^p near the lower limit of integration.

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Homework Statement


Find the values of p for which the integral converges and evaluate the integral for those values of p.

∫ 0->1 1/(x^p) dx

Homework Equations


None.


The Attempt at a Solution



First thought:

Since we must evaluate 0 to 1, 1/0 is undefined so maybe 1/ (0^0) = 1.
I don't think this is correct.

Second thought:
If the first thought didnt work, how about lim x->0.
Test:
p≥1
1/ (0.000000000000000000...01)^1
≈ ∞

0>p>1
1/ (0.000000000000000000...01)^.5
≈ ∞

p<0
1/ (0.000000000000000000...01)^-1
≈ 0

If this test is true, I am not sure how to evaluate for 1.

4. Answer in the back of the testbook
p<1 , 1/(1-p)

From the answer, I am not sure how p<1 would work (but I do understand p<0).
 
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You are making a test to guess if f(x)=1/x^p diverges at x=0. That doesn't tell you the integral necessarily diverges. Take your p=1/2 case. What the antiderivative of 1/x^(1/2)?
 

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