Question

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

Answer 1

\[\int\limits_{0}^{1}{1\over x^p}dx\]

Answer 2

\[ p<1\\
\int_0^1 \dfrac {dx}{x^p}= \left[ \dfrac { x^{1-p} } {1-p} \right]_0^1=
\int_0^1 \dfrac {dx}{x^p}= \left[ \dfrac { x^{1-p} } {1-p} \right]_0^1=\dfrac
1{1-p}-0 =\dfrac 1 {1-p}
\]

Answer 3

why is p<1?

Answer 4

For p >1, the integral diverges, since 1-p<0 and this makes
\[ x^{1-p}\] goes to \[+\infty\] as x tends to zero.

Answer 5

ok, HUH?

Answer 6

p< 1 for the integral to to converge. If p=1 the integral would be ln(x) wich goes to + Infinity when x goes to zero.

Answer 7

I get how you did the integral, somewhat, but not why p<1.

Answer 8

Try it for p=2 and see whay things do not work out.

Answer 9

-1-infinity

Answer 10

therefore it diverges
but how do you know that it will diverge without testing it?

Answer 11

\[
\int_c^1 \dfrac {dx}{x^p}= \left[ \dfrac { x^{1-p} } {1-p} \right]_c^1
=\dfrac 1 {1-p}-\dfrac { c^{1-p} } {1-p}
\]
You have to find the limit f=of the above for c going to zero.
It diverges if p>= 1 and converges if p<1

Answer 12

I think I understand your reasoning. One just needs to set p=1 first to get a grip on the "situation/question". Either than or just have a really keen understanding.

Also, just a little tidbit on what you said earlier. I think ln(x) approaches - infinity when x goes to 0, not positive infinity.