Denote $f(x)=\int_x^{x+1}\cos t^2 {\rm d}t.$Prove $\lim\limits_{x \to +\infty}f(x)=0.$

Question

Problem

Denote $$f(x)=\int_x^{x+1}\cos t^2 {\rm d}t.$$Prove $\lim\limits_{x \to +\infty}f(x)=0.$

Proof

Assume $x>0$. Making a substitution $t=\sqrt{u}$，we have ${\rm d}t=\dfrac{1}{2\sqrt{u}}{\rm d}u.$ Therefore, \begin{align*} f(x)&=\int_x^{x+1}\cos t^2 {\rm d}t\\ &=\int_{x^2}^{(x+1)^2} \frac{\cos u}{2\sqrt{u}}{\rm d}u\\ &=\frac{1}{2\sqrt{\xi}}\int_{x^2}^{(x+1)^2}\cos u{\rm d}u\\ &=\frac{\sin(x+1)^2-\sin x^2}{2\sqrt{\xi}}, \end{align*} where $x^2 \leq \xi\leq (x+1)^2$. Further, $$0\leq |f(x)|\leq \frac{|\sin(x+1)^2|+|\sin x^2|}{2\sqrt{\xi}}\leq \frac{1}{\sqrt{\xi}}\leq \frac{1}{x}\to 0(x \to +\infty),$$ which implies $f(x)\to 0(x \to +\infty)$, according to the squeeze theorem.

Answer 1

Alternative solution:
Denote $$g(x)=\int_x^\infty \cos t^2dt\ (x>0),$$ we can actually prove this definition is valid (integral converges) using substitution $t^2=u$ and Dirichlet's test.
According to the definition of improper integral: $$\lim_{y\to\infty}\int_x^y\cos t^2dt=\int_x^\infty\cos t^2dt,$$ we can deduce $$\lim_{x\to\infty}g(x)=\int_a^\infty\cos t^2dt-\lim_{x\to\infty}\int_a^x\cos t^2dt=\int_a^\infty\cos t^2dt-\int_a^\infty\cos t^2dt=0.$$ Therefore $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x+1)-g(x)=0$.

Answer 2

Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $\cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.

Another Proof

Likewise, assume $x>0$. Making a substitution $t=\sqrt{u}$，we have ${\rm d}t=\dfrac{1}{2\sqrt{u}}{\rm d}u.$ Therefore, \begin{align*} f(x)&=\int_x^{x+1}\cos t^2 {\rm d}t\\ &=\int_{x^2}^{(x+1)^2} \frac{\cos u}{2\sqrt{u}}{\rm d}u\\ &=\int_{x^2}^{(x+1)^2}\frac{1}{2\sqrt{u}}{\rm d}(\sin u)\\ &=\frac{\sin u}{2\sqrt{u}}\bigg|_{x^2}^{(x+1)^2}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}\sin u{\rm d}u\\ &=\frac{\sin(x+1)^2}{2(x+1)}-\frac{\sin x^2}{2x}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}\sin u{\rm d}u.\\ \end{align*} Thus \begin{align*} |f(x)|&\leq \frac{|\sin(x+1)^2|}{2(x+1)}+\frac{|\sin x^2|}{2x}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}|\sin u|{\rm d}u\\ &\leq \frac{1}{2(x+1)}+\frac{1}{2x}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}{\rm d}u\\ &=\frac{1}{2(x+1)}+\frac{1}{2x}+\frac{1}{4}\cdot \left[-\frac{2}{\sqrt{u}}\right]_{x^2}^{(x+1)^2}\\ &=\frac{1}{x}\to 0(x \to +\infty), \end{align*} which implies $|f(x)|\to 0$ according to the squeeze theorem. It follows that $f(x) \to 0$.