3
$\begingroup$

I have to determine if the improper integral $ \intop_{1}^{\infty}\sin\left(x^{p}\right)dx $

convergent/divergent for any $ 0<p \in \mathbb{R} $

Here's what Ive done :

We can substitute $ x^{p}=y $ and then we'll get

$ \intop_{1}^{\infty}\sin\left(x^{p}\right)dx=\intop_{1}^{\infty}y^{\frac{1-p}{p}}\sin\left(y\right) $

Thus, for $ p>1 $ the integral will converge by Dirichlet's theorem.

My intuition is that for $ p<1 $ the integral diverges.

I'll write it again, let $ \frac{1-p}{p}=\alpha $

How do we prove that $ \intop_{1}^{\infty}x^{\alpha}\sin\left(x\right)dx $ diverge for $ \alpha > 0 $ ?

I tried to show that $ x^{\alpha}\sin\left(x\right) $ will not follow Cauchy's condition but it got complicated.

Thanks in advance.

$\endgroup$
1
  • $\begingroup$ Some related content here $\endgroup$
    – Zakhurf
    Jul 18, 2020 at 15:11

1 Answer 1

0
$\begingroup$

$$ \int_{\pi}^{n\pi} x^{\alpha}\sin(x)dx=\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}x^{\alpha}\sin(x)dx=\sum_{k=1}^{n-1}(-1)^k\int_{0}^{\pi}(u+k\pi)^{\alpha}\sin(u)du $$ The integral $\int_{0}^{\pi}(u+k\pi)^{\alpha}\sin(u)du$ does not converge to $0$ as $k\rightarrow +\infty$ because, using $\sin(x)\geqslant\frac{2}{\pi}x$ for $x\in[0,\pi/2]$, $$ \int_{0}^{\pi}(u+k\pi)^{\alpha}\sin(u)du\geqslant\int_0^{\pi/2}(u+k\pi)^{\alpha}\sin(u)du\geqslant\frac{2}{\pi}\int_0^{\pi/2}(u+k\pi)^{\alpha}udu\geqslant\frac{2}{\pi}\int_0^{\pi/2}u^{\alpha+1}du\geqslant\frac{ (\pi/2)^{\alpha+1}}{\alpha+2} $$ Thus the integral $$\int_1^{\infty}x^{\alpha}\sin(x)dx$$ diverges for $\alpha>0$.

$\endgroup$
1
  • $\begingroup$ Nice user name ! $\endgroup$ Jul 18, 2020 at 14:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .