# Thread: Find values for which this integral diverges

1. ## 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?

2. ## 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:

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

3. ## Re: Find values for which this integral diverges

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

4. ## Re: Find values for which this integral diverges

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

5. ## 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?

6. ## 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

7. ## Re: Find values for which this integral diverges

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 $0$.

8. ## Re: Find values for which this integral diverges

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 $0$.