Find values for which this integral diverges

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June 4th 2012, 11:59 AM
infernalmich

Find values for which this integral diverges

I have the following question, I have to find the values of a for which the indefinite integral of x^-a from 1 to infinity diverge.

Substitution h for infinity I get: limit h->0 (1/(a-1)-h1^-y/(a-1))

So it diverges only for a=1 because there the limit is not defined! Is this correct?
June 4th 2012, 12:30 PM
FernandoRevilla

Re: Find values for which this integral diverges

If $a=1$ then, $\int_0^b\dfrac{dx}{x}=\log b$ . If $b\to +\infty$ , then $\log b\to +\infty$ and the integral is divergent. If $a\neq 1$ then, $\int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}$ . Easily proved:

$\int_0^{+\infty}\dfrac{dx}{x^a}=\begin{Bmatrix} +\infty & \mbox{ if }& a<1\\\dfrac{1}{a-1}& \mbox{if}& a>1\end{matrix}$
June 4th 2012, 12:45 PM
infernalmich

Re: Find values for which this integral diverges

Thank you! I didn't thought about writing x^-a as 1/x^a
June 5th 2012, 12:46 AM
FernandoRevilla

Re: Find values for which this integral diverges

Quote:

Originally Posted by FernandoRevilla

If $a\neq 1$ then, $\int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}$ .

Of course I meant $\int_0^b\dfrac{dx}{x^a}$ instead of $\int_0^b\dfrac{dx}{x^2}$ .
June 17th 2012, 07:52 AM
infernalmich

Re: Find values for which this integral diverges

If I do the same exercise with the Integral from 0 to infinity, it diverges for every "a" - is this true?
June 17th 2012, 08:28 AM
infernalmich

Re: Find values for which this integral diverges

What a satisfaction, I found the solution... I was wrong.

It converges for a<1 and diverges for a >1
June 20th 2012, 11:47 PM
FernandoRevilla

Re: Find values for which this integral diverges

Quote:

Originally Posted by FernandoRevilla

If $a=1$ then, $\int_0^b\dfrac{dx}{x}=\log b$ . If $b\to +\infty$ , then $\log b\to +\infty$ and the integral is divergent. If $a\neq 1$ then, $\int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}$ . Easily proved:

$\int_0^{+\infty}\dfrac{dx}{x^a}=\begin{Bmatrix} +\infty & \mbox{ if }& a<1\\\dfrac{1}{a-1}& \mbox{if}& a>1\end{matrix}$

Of course the lower limit is $1$ instead of $0$ .
June 20th 2012, 11:48 PM
FernandoRevilla

Re: Find values for which this integral diverges

Quote:

Originally Posted by FernandoRevilla

Of course I meant $\int_0^b\dfrac{dx}{x^a}$ instead of $\int_0^b\dfrac{dx}{x^2}$ .

Of course (again) the lower limit is $1$ instead of $0$ .