Can the improper integral involving the sine function and a polynomial converge?

[Ackbach]

Here is this week's POTW:

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Show that the improper integral
\[ \lim_{B\to\infty}\int_{0}^B \sin(x) \sin\left(x^2\right) \, dx\]
converges.

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[Ackbach]

Re: Problem Of The Week # 262 - May 08, 2017

This was Problem A-4 in the 2000 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows:

Spoiler

Let $A<B$. Then (integrating by parts) $$ \begin{aligned} \int_A^B \sin x \sin (x^2)\,dx &= \int_A^B \frac {\sin x}{2x} (2x\sin (x^2)) \,dx \\ &= \Bigl[ \frac {\sin x}{2x}(-\cos (x^2)) \Bigr]_A^B + \int_A^B \frac{x\cos x - \sin x}{2x^2}\cos (x^2)\,dx \\ &= \frac{-\sin B\cos (B^2)}{2B} + \frac{\sin A\cos (A^2)}{2A} + \int_A^B \frac{x\cos x - \sin x}{2x^2}\cos (x^2)\,dx. \end{aligned}$$ It follows from the triangle inequality that $$ \left| \int_A^B \sin x \sin (x^2)\,dx \right| \leqslant \frac1{2B} + \frac1{2A} + \left|\int_A^B \frac{x\cos x - \sin x}{2x^2}\cos (x^2)\,dx \right|.$$ Now integrate by parts again, to get $$ \begin{aligned} \int_A^B \frac{x\cos x - \sin x}{2x^2}\cos (x^2)\,dx &= \int_A^B \frac{x\cos x - \sin x}{4x^3}(2x\cos (x^2))\,dx \\ &= \Bigl[ \frac{x\cos x - \sin x}{4x^3}\sin(x^2)\Bigr]_A^B - \int_A^B \frac{(-x\sin x)x^3 - 3x^2(x\cos x - \sin x)}{4x^6}\sin(x^2)\,dx \\ &= \frac{B\cos B - \sin B}{4B^3} - \frac{A\cos A - \sin A}{4A^3} - \int_A^B \frac{-x^2\sin x - 3(x\cos x - \sin x)}{4x^4}\sin(x^2)\,dx. \end{aligned}$$ From the triangle inequality, $$ \begin{aligned} \left| \int_A^B \frac{x\cos x - \sin x}{2x^2}\cos (x^2)\,dx \right| &\leqslant \frac1{2B^2} + \frac1{2A^2} + \left|\int_A^B \frac{-x^2\sin x - 3(x\cos x - \sin x)}{4x^4}\sin(x^2)\,dx \right| \\ &\leqslant \frac1{2B^2} + \frac1{2A^2} + \int_A^B \left|\frac{-x^2\sin x - 3(x\cos x - \sin x)}{4x^4}\sin(x^2)\right |\,dx \\ &\leqslant \frac1{2B^2} + \frac1{2A^2} + \int_A^B \frac{8x^2}{4x^4}\,dx \\ &= \frac1{2B^2} + \frac1{2A^2} +\left[ -2x^{-1}\right]_A^B \leqslant \frac1{2B^2} + \frac1{2A^2} + \frac1B + \frac1A .\end{aligned} $$ Put that all together to see that $$\left|\int_A^B \sin x \sin (x^2)\,dx \right| \leqslant \frac1{2B} + \frac1{2A} + \frac1{2B^2} + \frac1{2A^2} + \frac1{B} + \frac1{A} \to0 \quad \text{as }\ A,B \to\infty.$$ It follows from the Cauchy criterion that $$\lim_{B\to\infty}\int_0^B \sin x \sin (x^2)\,dx$$ converges.