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I'm trying to understand the $Dirichelet \ Integral \ test$

I see in many places that the integral is like this: $$\int_{1}^{\infty}\frac{1}{x^k}dx$$

and if $k<=1$ it's $divergent$ otherwise it's $convergent$.

My question is about the lower limit. Can I use this test only on integrals starting at $1$ or it is valid for any given number?

Something like

$$\int_{a}^{\infty}\frac{1}{x^k}dx$$

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If $k>1$ then $$\int_1^a\frac{1}{x^k}dx$$ is finite for any $a>0$, we have that $$ \int_1^\infty\frac{1}{x^k}dx= \int_1^a\frac{1}{x^k}dx+\int_a^\infty\frac{1}{x^k}dx, $$ so $$ \int_1^\infty\frac{1}{x^k}dx\quad\text{and}\quad\int_a^\infty\frac{1}{x^k}dx $$ converge or diverge simultaneously (if $a>0$).

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Tanks for your answer –  Favolas Jan 25 '13 at 14:05
    
Just a quick question. For type II improper integrals can I assume that the limits in $\int_{0}^{1}\frac{1}{x^k}dx$ can be any kind? –  Favolas Jan 25 '13 at 15:31
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