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Can I use Dirichlet's test?

I know that$\displaystyle\int_{a}^{b}\sin x\le 1$.

Also, $\displaystyle\frac{\ln x}{x^2\sqrt{x^2-4}}$ is monotonic.

Also, $\displaystyle\lim_{x\to \infty} \frac{\ln x}{x^2\sqrt{x^2-4}} =0$

Can then conclude that the series converges?

Yuriy S
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newhere
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4 Answers4

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Yes, you can apply the Dirichlet test, since $ x \mapsto \dfrac{\ln x}{x^2\sqrt{x^2-4}}$is monotonic decreasing to $0$ over $[3, \infty)$ and $ \left|\int_3^b \sin x \:dx\right|<2. $

Community
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Olivier Oloa
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  • How do I prove that sin is bounded by $2$? I need to evalute $\int_{3}^{\infty}sinx dx$? – newhere Aug 04 '16 at 13:22
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    @newhere you have $\left| -\int_{3}^b \sin x dx \right|=\left| [-\cos x]_3^b \right|\le 2$. The upper bound is not $\infty$, please see the link I gave in my answer. – Olivier Oloa Aug 04 '16 at 13:24
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For large enough $x$, the integrand behaves like $\dfrac{\sin x\log x}{x^3}$.

Show that the integral of the modulus of this quantity, converges. You are done.

Landon Carter
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Actually , you do not need the Dirichlet test, because the integral is absolutely convergent: for $x\geq 3$, $$0\leq\frac{|\sin x| \cdot \ln x}{x^2\sqrt{x^2-4}}\leq \frac{1\cdot x}{x^2\sqrt{x^2-4}}\sim\frac{1}{x^2}$$ and $\int_3^{+\infty} \frac{dx}{x^2}$ is convergent.

Robert Z
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  • if the integral was just $\int_{2}^{\infty} \frac{dx}{x^2\sqrt{x^2-4}}$ could I say that it like $\frac{1}{x^2}$? – newhere Aug 04 '16 at 13:36
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    @newhere. You have better than that because $\frac{1}{x^2\sqrt{x^2-4}}\sim \frac{1}{x^3}$ as $x\to+\infty$. Moreover, near $2^+$ we have $\frac{1}{x^2\sqrt{x^2-4}}\sim \frac{1}{8\sqrt{x-2}}$. Hence your improper integral is convergent. – Robert Z Aug 04 '16 at 13:49
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For $x\geq3$, you have $\ln (x)/\sqrt {x^2-4}<1$, so $$ \int_3^\infty\left|\frac {\sin x\ \ln x}{x^2\,\sqrt {x^2-4 }}\right|\,dx = \int_3^\infty\frac {|\sin x|\ \ln x}{x^2\,\sqrt {x^2-4 }}\,dx \leq\int_3^\infty\frac {1}{x^2}\,dx <\infty $$ and the integral converges absolutely.

Martin Argerami
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