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This problem comes from a discussion with one of my friends:

Prove that: $$\displaystyle \lim_{n \to \infty}\int_{1}^{n}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \lim_{n \to \infty}\int_{1}^{n}{\sin(x^2)}\,{\mathrm dx}$$

I wonder if there is a reasonable way to somehow solve such a problem. Thanks.

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    |sin($x)|≤1$ can help?2012-05-26
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    @Babak Sorouh: i'm not sure how i may use it in a useful way. If you have a whole way to go, pls let me see it.2012-05-26
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    These are Fresnel integrals and could be evaluated in this way very easily. Is this enough an answer for you or you need a simpler way to get it?2012-05-26
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    @Jon: if you could come up with a simple proof it would be great.2012-05-26
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    @Chris: Indeed, one can give these integrals in closed form through Fresnel functions as $S\left(\sqrt{\frac{2}{\pi}}\right)\sim 0.247558$. So, on the lhs you will have $\frac{1}{2}\sqrt{\frac{\pi}{2}}\left(1-2S\left(\sqrt{\frac{2}{\pi}}\right)\right)\sim 0.316389$ and so on.2012-05-26
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    @Jon Please post this as an answer.2012-05-27

2 Answers 2

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The idea goes as follows. Consider your inequality $$ \displaystyle \lim_{n \to \infty}\int_{1}^{n}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \lim_{n \to \infty}\int_{1}^{n}{\sin(x^2)}\,{\mathrm dx} $$ but both integrals are meaningful in the given limit being standard values of Fresnel functions. So, you have $$ \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \int_{1}^{\infty}{\sin(x^2)}\,{\mathrm dx}. $$ The rhs can be immediately evaluated to give $$ \int_{1}^{\infty}{\sin(x^2)}\,{\mathrm dx}=\frac{1}{2}\sqrt{\frac{\pi}{2}}\left[1-2S\left(\sqrt{2}{\pi}\right)\right]. $$ I have introduced here a Fresnel function. These are defined in the following way $$ S(x)=\int_0^x dt\sin\left(\frac{\pi}{2}t^2\right)\qquad C(x)=\int_0^x dt\cos\left(\frac{\pi}{2}t^2\right) $$ and $\frac{\pi}{2}$ factor that you see in the integrals is the reason why your integral is evaluated at that value for $S$. For your case is $S\left(\sqrt{2}{\pi}\right)\sim 0.247558$ and so the rhs evaluates to 0.316389.

On the lhs you have to do some more work. You will have $$ \sin(x)\sin(x^2)=\frac{1}{2}\left[\cos(x-x^2)-\cos(x+x^2)\right] $$ and so you have to evaluate $$ \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}=\frac{1}{2}\int_{1}^{\infty}\left[\cos(x-x^2)-\cos(x+x^2)\right]\,{\mathrm dx}. $$ In order to evaluate these integrals we firstly note that $x^2+x=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}$ and $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ and so $$ \frac{1}{2}\int_{1}^{\infty}\left[\cos(x-x^2)-\cos(x+x^2)\right]\,{\mathrm dx}= \frac{1}{2}\int_{\frac{1}{2}}^{\infty}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}-\frac{1}{2}\int_{\frac{3}{2}}^{\infty}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy} $$ after an obvious change of variables in the integrals. But we recognize immediately that $$ \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}= \frac{1}{2}\int_{\frac{1}{2}}^{\frac{3}{2}}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy} $$ and expanding the cosine we realize that $$ \int_{\frac{1}{2}}^{\frac{3}{2}}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}= \int_{\frac{1}{2}}^{\frac{3}{2}}\cos(y^2)\cos\left(\frac{1}{4}\right)\,{\mathrm dy}+\int_{\frac{1}{2}}^{\frac{3}{2}}\sin(y^2)\sin\left(\frac{1}{4}\right)\,{\mathrm dy} $$ and we are back to Fresnel functions. Already at this stage, you should be able to realize that the inequality holds as you are evaluating Frensel functions on a smaller interval and these are also multiplied by factors that are lesser than unity. But we can go on with numerical evaluation obtaining 0.286035 that is lower than 0.316389 as it should.

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    very nice solution! Thanks!2012-05-27
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    it's a bit over my knowledge, but this is a good opportunity to brush up my knowledge on Fresnel integrals.2012-05-27
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    @Chris: You are welcome.2012-05-27
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May be there is none, general at least, since if to change the lower limits of integration from $1$ to $a$ the inequality wouldn't hold for all $a$. Here is a plot of the limits for different values of $a$, lhs in red and rhs in blue:

enter image description here

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    Of course, this will depend on the way Fresnel functions will evaluate at different $a$. Why to hide the objects we are talking about?2012-05-26
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    @Jon the rhs can be expressed through a Fresnel integral, but not the lhs. Or am I missing something?2012-05-26
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    Yes, you can. The trick is the same of when you do the Fourier transform of a Gaussian. You will complete the square and you will get again a Gaussian. So, if you consider the more general integral $\int dz e^{iz+iz^2}$ you can do the trick.2012-05-26