Prove $|\int_{n}^{n+p} \sin (x^2)dx|\leq 1/n$ where $p>0$.

Question

Prove $$\left|\int_{n}^{n+p} \sin (x^2)dx\right|\leq \frac{1}{n}$$ where $p>0$.

Maybe, we can improve namely enhance to prove $$\left|\int_{n}^{+\infty} \sin (x^2)dx\right|\leq \frac{1}{n},$$ which is hold?

Answer 1

You can say that: $$\int_n^\infty\sin(x^2)dx=\Im\int_n^\infty e^{ix^2}dx=\Im[I]$$ Now we will square $I$ and make a change of variable to get: $$I^2=\int_n^\infty\int_n^\infty e^{i(x^2+y^2)}dxdy=\int_{-\pi}^\pi\int_{\sqrt{2}n}^\infty re^{ir^2}drd\theta=\frac12\int_{-\pi}^\pi\int_{2n^2}^\infty e^{iu}dud\theta=\frac{1}{i}\left[e^{iu}\right]_{2n^2}^\infty$$ However this upper limit is a problem as it is indeterminate, all we know is that: $$|\lim_{u\to\infty}e^{iu}|=1$$ However based upon assumptions made of this value you could evaulate the integral. If we assume it has a value of $0$ then we can say: $$I^2=\frac{-e^{2in^2}}{i}$$

Answer 2

I have found a proof as follows. Please correct me if I'm wrong.

Proof

Make a substitution $x^2=:t\geq 0$. Then $x=\sqrt{t}$，$dx=\dfrac{1}{2\sqrt{t}}dt$. Therefore $$\int_{n}^{n+p}\sin x^2 dx=\int_{n^2}^{(n+p)^2}\frac{\sin t}{2\sqrt{t}}dt=\frac{1}{2}\int_{n^2}^{(n+p)^2}\frac{\sin t}{\sqrt{t}}dt.$$ Notice that $\dfrac{1}{\sqrt{t}}$ is decreasing over $[n^2,(n+p)^2]$ and $\dfrac{1}{\sqrt{t}}\geq 0$. As per the second integral mean value theorem, there exists $ \xi \in [n^2,(n+p)^2]$ such that \begin{align*} \:\left|\int_{n}^{n+p}\sin x^2 dx\right|&=\frac{1}{2}\left|\int_{n^2}^{(n+p)^2}\frac{\sin t}{\sqrt{t}}dt\right|=\frac{1}{2}\left|\frac{1}{n}\int_{n^2}^{\xi}\sin t dt\right|\\ &=\frac{1}{2n}\left|\int_{n^2}^{\xi}\sin t dt\right|=\frac{1}{2n} \cdot |\cos \xi-\cos n^2|\\ &\leq \frac{1}{2n}\cdot( |\cos \xi|+|\cos n^2|)\\ &\leq \frac{1}{2n} \cdot (1+1)\\ &=\frac{1}{n}, \end{align*} which is desired.