For which $\alpha \in \mathbb{R}$ this integral converges?

Question

Consider we have $$\int_0^1 \frac{\sin(\frac{1}{x})}{(\sqrt{x} - x)^\alpha}dx$$

As I understand I have to use Dirichlet's test, but is there all theorem's conditions are satisfied?

Maybe we have to make substitution $t = \frac{1}{x}$ and it helps? I got some troubles with this and maybe someone can help me.

Dirichlet's test is for series, not for improper integrals. — José Carlos Santos, May 22 '21 at 09:31
@JoséCarlosSantos as I know it also works for integrals with similar conditions. Here's a post about it — Someone, May 22 '21 at 09:35
@José Carlos Santos there is a Dirichlet's test for improper integral, see https://en.m.wikipedia.org/wiki/Dirichlet%27s_test#Improper_integrals — William, May 22 '21 at 09:35

William · Answer 1 · 2021-05-22T09:58:47.820

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Are you sure $\alpha$ is any real number? Can you state the definition of Dirichlet's Test that you're using? The one I know is—

Dirichlet's Test: If $g(x)$ is a bounded monotonic function for $x≥a$ for some $a \in \mathbb{R}$ such that $\lim\limits_{x \to ∞} g(x)=0$ and $f$ is a function such that $\displaystyle \left| \int_a^x f(t) \, \mathrm{d}t \right|$ is bounded for $x≥a$ then $\displaystyle \int_a^∞ f(x)g(x) \, \mathrm{d}x$ is convergent.

I think if you this test, it will not work for all real values of $\alpha$

edited May 22 '21 at 09:58

answered May 22 '21 at 09:52

William

5,151

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Thank you. Test really works for all $\alpha > 0$. But can we say something about other values of $\alpha$ maybe without a test using different ideas? – Someone May 22 '21 at 10:00
I suppose you can that the Dirichlet's test is inconclusive? – William May 22 '21 at 10:05
That's reason why I'm asking other ways to find $\alpha$ – Someone May 22 '21 at 10:16

For which $\alpha \in \mathbb{R}$ this integral converges?

1 Answers1