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I have to show that $f(x)=\sin(x^2)$ is integrable on $[1, \infty[$. This is French terminology, so "intégrable" specifically means that the integral of $|f|$ exists.

The only method I know is to compare it to functions of the form $\frac{1}{x^\alpha}$, but it's not eventually smaller or larger than any of these. I can't imagine how it could be asymptotically equivalent to anything useful either, seeing as it oscillates like crazy.

Jack M
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    I wouldn't be so sure it's absolute value is integrable, but maybe you can make an argument on the width of the base of your peaks, which have all height $1$. You can also make a change of variables $u=x^2$ to get $$\dfrac 1 2 \int_1^\infty |\sin u|\frac{du}{\sqrt u}$$ – Pedro Oct 10 '13 at 19:03
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    With that last approach you can compare to $\dfrac 1 4 \dfrac{1}{\sqrt u}$ in a rather tricky way. This last function cuts the lumps of your function pretty halfway, then you can argue that the top curved triangle can be used to fill up (with a little extra area left) the "hollow" pointy triangle below. But $$\int_1^\infty \frac{du}{\sqrt u}$$ diverges, so your integral diverges too. – Pedro Oct 10 '13 at 19:08
  • This question is a hit. – user64494 Oct 10 '13 at 19:21
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    @PedroTamaroff: the Lebesgue integral does not converge since the integral of the absolute value diverges. However, the Riemann integral for $[0,\infty)$ is defined as the limit as the upper bound tends to infinity. – robjohn Oct 10 '13 at 19:21
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    @robjohn I know that Rob! =D – Pedro Oct 10 '13 at 22:14

2 Answers2

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If you mean that $$ \lim_{N\to\infty}\int_0^N\sin(x^2)\,\mathrm{d}x $$ exists, then change variables $x\mapsto\sqrt{x}$ and integrate by parts: $$ \begin{align} &\lim_{N\to\infty}\int_0^N\sin(x^2)\,\mathrm{d}x\\ &=\int_0^1\sin(x^2)\,\mathrm{d}x +\lim_{N\to\infty}\frac12\int_1^{N^2}\frac{\sin(x)}{x^{1/2}}\,\mathrm{d}x\\ &=\int_0^1\sin(x^2)\,\mathrm{d}x +\lim_{N\to\infty}\frac12\left[\frac{1-\cos(x)}{x^{1/2}}\right]_1^{N^2} +\lim_{N\to\infty}\frac14\int_1^{N^2}\frac{1-\cos(x)}{x^{3/2}}\,\mathrm{d}x\\ \end{align} $$ Now each piece has a limit as $N\to\infty$ since $$ \int_0^1\sin(x^2)\,\mathrm{d}x $$ is constant $$ \lim_{N\to\infty}\frac12\left[\frac{1-\cos(x)}{x^{1/2}}\right]_1^{N^2}=\frac{\cos(1)-1}2 $$ and $$ \left|\frac{1-\cos(x)}{x^{3/2}}\right|\le\frac2{x^{3/2}} $$ which is integrable over $[1,\infty]$ since $\frac32\gt1$.

robjohn
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  • Why $1-\cos(x)$ rather than just $\cos(x)$? – Jack M Oct 10 '13 at 19:25
  • @JackM Because it works. =) – Pedro Oct 10 '13 at 19:25
  • @PedroTamaroff Wouldn't $-\cos(x)$? – Jack M Oct 10 '13 at 19:29
  • @JackM Nope, since it is unbounded $x\to 0$ when divided by $x^\alpha$. – Pedro Oct 10 '13 at 19:29
  • @JackM: because I was going to use the same integration by parts over all of $[0,\infty)$, because $\frac{1-\cos(x)}{x^{3/2}}$ vanishes at $0$ and at $\infty$. However, if that was not obvious to anyone, the explanation would be more confusing that simply breaking the domain into $[0,1)$ and $[1,\infty)$. I changed the domain, but left the integration by parts. After changing the domain, $-\cos(x)$ would have worked, too. – robjohn Oct 10 '13 at 19:30
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Your function is not Lebesgue integrable. As I commented, the absolute value is not integrable. By $u=x^2$ we get $$\int_1^\infty \frac{|\sin u|}{2\sqrt u}du$$

Now, consider $f(u)=\frac{1}{4\sqrt u}$. Our integrand is red, this last function is blue. We know that $$\int_1^\infty \frac{du}{\sqrt u}$$ diverges. From the graph, we can see that $$\int_1^\infty \frac{|\sin u|}{2\sqrt u}du> \int_1^\infty \frac{1}{4\sqrt u}du$$ by comparing the size of the red and green triangles. Thus this integral diverges.

(I guess one can make this rigorous, but I think the argument is clear and convincing.)

enter image description here

ADD The intersection with the $x$-axis are the points $x=\pi k$, i.e. the zeroes of $\sin x$. Solving for the extrema (i.e. using derivatives) gives $\tan x=2x$. These points behave asymtotically like $x=\frac \pi 2+\pi k$. Finally, the intersection of the blue graph with the red graph is given by $|\sin x|=\dfrac 1 4$, thus I don't see it impossible to estimate accurately the areas of the triangles in question.

Pedro
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  • @TBrendle What do you mean? – Pedro Oct 10 '13 at 19:26
  • How can you compare the sizes of the triangles? For a start they're not really triangles as one of their sides is a hyperbola, and secondly their heights and widths don't seem trivial to calculate at all. – Jack M Oct 10 '13 at 19:32
  • @JackM I really do mean to use triangles, not the curved ones. The green one is still smaller than the curved one over the blue line, yet larger than the red ones, so evidently the size of the curved triangle will be larger than the red ones. And yes, I don't think calculation is trivial, but I thought it was worth sketching why the integral of the absolute value diverged. – Pedro Oct 10 '13 at 19:36
  • (Actually, finding the coordinates of the triangle's vertices is not difficult at all). – Pedro Oct 10 '13 at 19:38
  • @TBrendle I know. If you want I'll add them, but I cannot do so now. – Pedro Oct 10 '13 at 19:39
  • I said the integral of the absolute value diverges, not that the integral diverges. – Pedro Oct 10 '13 at 19:40