Find values for which this integral diverges

**infernalmich** · June 4th 2012, 11:59 AM

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?

**FernandoRevilla** · June 4th 2012, 12:30 PM

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:

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

**infernalmich** · June 4th 2012, 12:45 PM

Thank you! I didn't thought about writing x^-a as 1/x^a

**FernandoRevilla** · June 5th 2012, 12:46 AM

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 $\int_0^b\dfrac{dx}{x^2}$ .

**infernalmich** · June 17th 2012, 07:52 AM

If I do the same exercise with the Integral from 0 to infinity, it diverges for every "a" - is this true?

**infernalmich** · June 17th 2012, 08:28 AM

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

It converges for a<1 and diverges for a >1

**FernandoRevilla** · June 20th 2012, 11:47 PM

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 $0$ .

**FernandoRevilla** · June 20th 2012, 11:48 PM

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 $0$ .

Thread: Find values for which this integral diverges

LinkBack

Thread Tools

Display

Find values for which this integral diverges

Re: Find values for which this integral diverges

Re: Find values for which this integral diverges

Re: Find values for which this integral diverges

Re: Find values for which this integral diverges

Re: Find values for which this integral diverges

Re: Find values for which this integral diverges

Re: Find values for which this integral diverges

Similar Math Help Forum Discussions

Converges or diverges - find its limit

trying to find if a particular series converges or diverges

How to find whether this series converges or diverges

double integral - converges or diverges

Find integral values

Search Tags