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I know how to solve and prove that

$$\int_{0}^{\infty} \frac{\sin(x)}{x^\alpha} \, dx$$ converge for $ 0 < \alpha < 2 $ with regular tests and integration by parts.

But with the Dirichlet test, I just see that I have one function which is monotonic decreasing to $0$ for any given $\alpha$ and the integral on the $\sin(x)$ is bounded for $ [0,\infty]$ because so it should converge. so it should converge for any $\alpha$.

Clearly I don't understand the Dirichlet test, but I've read its definition many times and still I don't understand where I got it wrong.

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Don't abuse the display style! Use them appropriately. Thanks to Nunoxic for his substantial edit and I have as well edited the post substantially. There is a gap: because so. I don't think they ever go together. –  user21436 Feb 14 '12 at 16:37
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Does the behaviour of $1/x^\alpha$ near 0 play well with Dirichlet? –  David Mitra Feb 14 '12 at 16:38
    
thanks. I just dont know how to add Latex in the same line, everytime i am inserting Latex code its just jumps to the next line without any \n newline character. and regarding 1/x^\alpha - isnt it monotonic decreasing for any given alpha? or that i am looking it wrong and because its not "proper" integrable in [0,infinity] i cant use it? –  YNWA Feb 14 '12 at 16:42

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Dirichlet Test

Let $f$, $g :[a,\infty)\rightarrow\Bbb R $ be such that:

1) $f$ is decreasing and $f(x)\rightarrow0$ as $x\rightarrow\infty$.

2) $g$ is continuous and there is an $M$ such that $\Bigl|\int_a^x g(t)\,dt\Bigr|\le M$ for all $x>a$.

Then $\int_a^\infty f(t)g(t)\,dt$ converges.


Now, is $f(x)={1\over x^\alpha}$ defined on all of $[0,\infty)$? Could you even define an extension of $f$ to $[0,\infty)$ so that it remains monotone?

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you' the man...thank you! that part seemed to have vanished from my notebook –  YNWA Feb 14 '12 at 18:47

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