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Show that for any $\alpha>1$ the function $$ g_{\alpha}(x):=\sin(x^{\alpha}), x\geq 0 $$ is improperly Riemann-integrable but not Lebesgue-integrable on $\mathbb{R}_+$.

Could you please give me a hint how to show that?

Concerning the Riemann-integrability I have to show that $$ \lim\limits_{R\to\infty}\int\limits_0^R\sin(x^{\alpha})\, dx<\infty, $$ but I do not know how to show that.

Concerning the Lebesgue-integrability I have to show that $$ \int\limits_0^{\infty}\lvert\sin(x^{\alpha})\rvert\, dx=\infty, $$ but again I have no idea how to do so.

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3 Answers 3

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First substitute $u= x^{\alpha}$ into the integral to see your problem reduces to asking the same question about the integral $\displaystyle \int^{\infty}_0 \dfrac{\sin x}{x^a} dx$ for $a\in (0,1).$

To show that is Riemann integrable, we consider the integral over $(0,1)$ and $[1,\infty)$ separately. To show the integral over $[1,\infty)$ exists, integrate by parts. You'll have

$$ \int^R_1 \frac{\sin x}{x^a} dx = -x^{-a} \cos x \mid^R_1 - a \int^R_1 \frac{\cos x}{x^{a+1}} dx$$

and it should be easy to finish from there.

To see $\displaystyle \int^1_0 \frac{\sin x}{x^a} dx$ exists note that $\sin x \sim x$ and $\displaystyle \int^1_0 \frac{x}{x^a} dx$ is finite.

To show it is not Lebesgue integrable, it suffices to show $$\int^{\infty}_0 \frac{|\sin x|}{x^a} dx = \sum_{k=0}^{\infty} (-1)^k \int^{(k+1)\pi}_{k\pi} \frac{\sin x}{x^a} dx = \sum_{k=0} \int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx$$ diverges, which is not hard with a basic estimate on the last integral. Note that the integral over $[0,\pi]$ is certainly greater than the integral over $[\pi/4,3\pi/4]$ and on that interval we have $\sin x \geq \frac{1}{\sqrt{2}}$ so

$$\int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{1}{( x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} \frac{1}{( 3\pi/4 + k\pi)^a}= \frac{1}{2\sqrt{2}\pi^{a-1}}\frac{1}{(k+3/4)^a} .$$

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  • $\begingroup$ When I substitute $s=x^{\alpha}$, I get the integral $\int_0^{\infty}\frac{\sin(s)}{\alpha x^{\alpha-1}}\, ds$. Is that right?` $\endgroup$
    – math12
    Aug 17, 2013 at 12:23
  • $\begingroup$ @math12 Yes it is, but now that you are integrating with respect to $s$ you should replace that $x$ in the denominator with a function of $s$ instead. $\endgroup$ Aug 17, 2013 at 12:27
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    $\begingroup$ Neither term in the RHS of the displayed identity in the "Riemann" part of the post, exists. $\endgroup$
    – Did
    Aug 17, 2013 at 12:50
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    $\begingroup$ Sorry, you have to deal with $(0,1)$ and $[1,\infty)$ separately. For $[1,\infty)$ the method I wrote above works, I will edit my answer now to address the other part. $\endgroup$ Aug 17, 2013 at 12:55
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    $\begingroup$ @math12 Not quite. You show that the integral on the RHS converges absolutely with that estimate you wrote, and then that means it converges. $\endgroup$ Aug 17, 2013 at 14:16
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I felt the counterexamples mentioned before are complicated and not straightforward to justify/remember. Instead, we can consider the following relatively simpler counterexample.

Consider the function $f(x)=\chi_{[n,n+1]}(x)\left(\frac{(-1)^n}{n}\right).$ $f$ is improper Riemann integrable and $\int\limits_{\mathbb{R}}f(x)dx=\sum\limits_{n \in \mathbb{N}}\frac{(-1)^n}{n} < \infty,$ but not Lebesgue integrable as $\int\limits_{\mathbb{R}} \left|f(x)\right|dx=\sum\limits_{n \in \mathbb{N}}\frac{1}{n}=\infty.$

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For the first one, reduce to show that $\int_1^\infty\sin(t^\alpha)dt$ is convergent. A substitution like $s=t^\alpha$ and an integration by part will give what we want.

For the second one, use again the same trick, then the inequality $\sin^2s\leqslant|\sin s|$. A trigonometric identity will do the job.

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