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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.

Willie Wong
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3 Answers3

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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

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

Ragib Zaman
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  • When I substitute $s=x^{\alpha}$, I get the integral $\int_0^{\infty}\frac{\sin(s)}{\alpha x^{\alpha-1}}, ds$. Is that right?` –  Aug 17 '13 at 12:23
  • @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. – Ragib Zaman Aug 17 '13 at 12:27
  • Thanks. Then I have the integral $\int_0^{\infty}\frac{\sin(s)}{\alpha s^{1-\frac{1}{\alpha}}}, ds$ . And now I can out the factor $\frac{1}{\alpha}$ in front of the integral and define $\beta:=1-\frac{1}{\alpha}$. Then I have your equivalent problem. –  Aug 17 '13 at 12:34
  • @math12 That's correct. Now just note that the factor of $1/\alpha$ in the integral does not really matter in answering the question so we can ignore it, and that $1-1/\alpha$ exponent in the denominator lies in the range $(0,1),$ so the problem really does reduce to what I said it does in the answer. – Ragib Zaman Aug 17 '13 at 12:37
  • Thanks now I try the rest. –  Aug 17 '13 at 12:38
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    Neither term in the RHS of the displayed identity in the "Riemann" part of the post, exists. – Did Aug 17 '13 at 12:50
  • Yes then I get for the partly integration a term like $0^{-a}-\cos(R)R^{-a}$... hmm –  Aug 17 '13 at 12:52
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    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. – Ragib Zaman Aug 17 '13 at 12:55
  • What does the notation $\sin x\sim x$ mean? –  Aug 17 '13 at 13:24
  • @math12 If you don't know what it means, you can instead just use $x/2 \leq \sin x \leq x$ on $[0,1].$ – Ragib Zaman Aug 17 '13 at 13:29
  • In order to finish the $[1,\infty)$ "part" of the proof: Do I have to estimate the integral on the RHS by $\leq\int_1^{R}\frac{1}{s^{\alpha+1}}, ds=\frac{1}{\alpha}-\frac{1}{\alpha R^{\alpha}}$? –  Aug 17 '13 at 14:05
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    @math12 Not quite. You show that the integral on the RHS converges absolutely with that estimate you wrote, and then that means it converges. – Ragib Zaman Aug 17 '13 at 14:16
  • It is: $\int_1^R\frac{\cos(s)}{s^{\alpha+1}}, ds\leq\lvert\int_1^R\frac{\cos(s)}{s^{\alpha+1}}, ds\rvert\leq\int\limits_1^R\lvert\frac{\cos(s)}{s^{\alpha+1}}\rvert, ds\leq\int\limits_1^R\frac{1}{s^{\alpha+1}}, ds=\frac{1}{\alpha}-\frac{1}{\alpha \cdot R^{\alpha}}$ - - Right? –  Aug 17 '13 at 14:24
  • @math12 It is a little more subtle than that. We want to show an integral $I:= \int_X f(x) dx$ converges. You only establish $ I < \infty.$. Even bounding the other side as well ($-\infty < I < \infty$) does not ensure convergence- $I$ may oscillate between finite values just as $\int^{R}_0 \sin x dx$ does. To show convergence of our integral, We use the following fact: Say $\int_X f(x) dx$ converges absolutely if $\int_X |f(x)|dx$ converges. If an integral converges absolutely, it converges. – Ragib Zaman Aug 17 '13 at 14:42
  • So I only have to argue that the integral $\int_1^{\infty}\frac{1}{s^{\alpha+1}}, ds$ exists and from this it follows the existence of $\int\limits_1^{\infty}\frac{\cos(s)}{s^{\alpha+1}}, ds$? –  Aug 17 '13 at 15:02
  • @math12 Well, not just from that $\int^{\infty}_1 \frac{1}{s^{\alpha+1}} dx $ exists, but also by using the theorem I mentioned in my last comment. – Ragib Zaman Aug 17 '13 at 15:06
  • Unfortunately I do not see why convergence follows from absolute convergencr ;( –  Aug 17 '13 at 15:53
  • @math12 Read the first two pages of this. – Ragib Zaman Aug 17 '13 at 16:30
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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 $\int\limits_{\mathbb{R}} \left|f(x)\right|dx=\sum\limits_{n \in \mathbb{N}}\frac{1}{n}=\infty.$

Celestina
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0

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.

Davide Giraudo
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