# Find values for which this integral diverges

• Jun 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?
• Jun 4th 2012, 12:30 PM
FernandoRevilla
Re: Find values for which this integral diverges
If $\displaystyle a=1$ then, $\displaystyle \int_0^b\dfrac{dx}{x}=\log b$. If $\displaystyle b\to +\infty$, then $\displaystyle \log b\to +\infty$ and the integral is divergent. If $\displaystyle a\neq 1$ then, $\displaystyle \int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}$. Easily proved:

$\displaystyle \int_0^{+\infty}\dfrac{dx}{x^a}=\begin{Bmatrix} +\infty & \mbox{ if }& a<1\\\dfrac{1}{a-1}& \mbox{if}& a>1\end{matrix}$
• Jun 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/xa
• Jun 5th 2012, 12:46 AM
FernandoRevilla
Re: Find values for which this integral diverges
Quote:

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

Of course I meant $\displaystyle \int_0^b\dfrac{dx}{x^a}$ instead of $\displaystyle \int_0^b\dfrac{dx}{x^2}$ .
• Jun 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?
• Jun 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
• Jun 20th 2012, 11:47 PM
FernandoRevilla
Re: Find values for which this integral diverges
Quote:

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

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

Of course the lower limit is $\displaystyle 1$ instead of $\displaystyle 0$.
• Jun 20th 2012, 11:48 PM
FernandoRevilla
Re: Find values for which this integral diverges
Quote:

Originally Posted by FernandoRevilla
Of course I meant $\displaystyle \int_0^b\dfrac{dx}{x^a}$ instead of $\displaystyle \int_0^b\dfrac{dx}{x^2}$ .

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