Evaluate $\lim_{a\rightarrow \infty }\frac{1}{a}\int_{0}^{a}\sin(x)\sin(x^2)\,dx$

Calculate $$I=\lim_{a\rightarrow \infty }\frac{1}{a}\int_{0}^{a}\sin(x)\cdot \sin(x^2)\,dx$$

I tried to use the fact that $$\lim_{x\rightarrow \infty }\frac{1}{x}\int_{0}^{x}(f(t))\,dt=\frac{1}{T}\int_{0}^{T}f(t)\,dt$$ where $$T$$ is period of the function.

So in my case $$T=2\pi$$, right? So I have $$I=\lim_{a\rightarrow \infty }\frac{1}{2\pi}\int_{0}^{2\pi}\sin(x)\sin(x^2)\,dx=\lim_{a\rightarrow \infty }\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\sin(x^2)\,dx$$

Now I noted $$x=\pi-t$$ so $$I=\lim_{a\rightarrow \infty }\frac{1}{\pi}\int_{0}^{\pi}\sin(\pi-t)\sin((\pi-t)^2)\,dt=\lim_{a\rightarrow \infty }\frac{1}{\pi}\int_{0}^{\pi}\sin(t)\sin((\pi-t)^2)\,dt$$

How to continue? Is my method correct?

• The function is not periodic... – PierreCarre May 3 '19 at 11:13
• Then how to approach this exercise? – DaniVaja May 3 '19 at 11:13
• @DaniVja The integral can be explicitly computed using Fresnel integrals, but if you are able to show that $\int_0^a \sin x \sin (x^2) dx$ is bounded, you don't need to compute the integral. – PierreCarre May 3 '19 at 11:32

Note that $$\frac{d}{dx}\left(\frac{\sin(x)\cos(x^2)}{2x}\right)=\frac{\cos(x)\cos(x^2)}{2x}-\frac{\sin(x)\cos(x^2)}{2x^2}-\sin(x)\sin(x^2)$$ Hence, for $$a>1$$, $$\int_{1}^{a}\sin(x)\sin(x^2)\,dx=\left[-\frac{\sin(x)\cos(x^2)}{2x}\right]_1^a+\int_{1}^{a}\frac{\cos(x)\cos(x^2)}{2x}\,dx-\int_{1}^{a}\frac{\sin(x)\cos(x^2)}{2x^2}\,dx$$ and therefore, as $$a\to+\infty$$, $$\left|\frac{1}{a}\int_{1}^{a}\sin(x)\sin(x^2)\,dx\right|\leq \frac{1}{a}\left( \frac{1}{2}+\frac{1}{2a}+\int_{1}^{a}\frac{dx}{2x}+\int_{1}^{a}\frac{dx}{2x^2}\right)= \frac{1+\frac{\log(a)}{2}}{a}\to 0$$ Finally $$\lim_{a\rightarrow +\infty }\frac{\int_{0}^{a}\sin(x)\sin(x^2)\,dx}{a}=\lim_{a\rightarrow \infty }\frac{\int_{0}^{1}\sin(x)\sin(x^2)\,dx}{a}+\lim_{a\rightarrow \infty }\frac{\int_{1}^{a}\sin(x)\sin(x^2)\,dx}{a}=0.$$

Remark. The limit can be evaluated also by using a stronger result: the improper integral $$\int_{0}^{+\infty}\sin(x)\sin(x^2)\,dx$$ is convergent (see Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge).

You don't really need to compute the integral. If you show that $$\int_0^a \sin x \sin (x^2) \,dx$$ is bounded (which it is), the limit is zero, since you are multiplying a bounded function by an infinitesimal (as $$a \to \infty$$).

• Can you give a hint how to show that boundedness? Because that will IMO necessarily involve lots of integration inequalities, or is there a way to avoid that? – Ingix May 3 '19 at 11:47
• @Ingix I do not know a simple way of showing boundness. In the end it is not more complicated to show that the improper integral is convergent (there are posts with that proof). – PierreCarre May 3 '19 at 11:52

$$\sin(x^2)$$

is not periodic, so this approach is not working.

Generally,

$$\sin((x+2\pi)^2) = \sin(x^2+4\pi x +4\pi^2) \neq \sin(x^2).$$

Even when $$x \in \mathbb N$$, when the $$4\pi x$$-part is a multiple of the period of $$\sin(x)$$, the $$4\pi^2$$-part makes the sin-values different.

No other period exits. As $$x \to \infty$$, the difference in arguments between consecutive maxima (=$$1$$) and minima (=$$-1$$) becomes smaller and smaller, as $$x^2$$ increases in values as well as inclination.