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

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  • Thread starter Ackbach
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    2017
In summary, the sine function and a polynomial can converge in an improper integral if the integral is evaluated within the appropriate limits and the function is continuous within those limits. The integral must be evaluated within appropriate limits, and the function must be continuous and bounded within those limits. However, it is possible for the improper integral to diverge if the function is not continuous within the limits of integration, or if it does not approach a finite limit as the limits approach infinity. The degree of the polynomial does not solely determine the convergence of the improper integral, as the behavior of the sine function and the limits of integration also play a significant role. The sine function is often used as a test function to determine the convergence or divergence of an improper integral due to
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Ackbach
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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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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
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:

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 \(\displaystyle \lim_{B\to\infty}\int_0^B \sin x \sin (x^2)\,dx\) converges.
 

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

Can the sine function and a polynomial converge in an improper integral?

Yes, the sine function and a polynomial can converge in an improper integral if the integral is evaluated within the appropriate limits and the function is continuous within those limits.

What conditions must be met for the improper integral involving the sine function and a polynomial to converge?

The integral must be evaluated within appropriate limits, and the function must be continuous and bounded within those limits. Additionally, the function must approach a finite limit as the limits of integration approach infinity.

Is it possible for the improper integral involving the sine function and a polynomial to diverge?

Yes, it is possible for the improper integral to diverge if the function is not continuous within the limits of integration, or if it does not approach a finite limit as the limits approach infinity.

Can the convergence of the improper integral be determined by the degree of the polynomial?

No, the degree of the polynomial does not solely determine the convergence of the improper integral. The behavior of the sine function and the limits of integration also play a significant role in determining convergence.

What is the significance of the sine function in an improper integral involving a polynomial?

The sine function is often used as a test function to determine the convergence or divergence of an improper integral. Its oscillatory behavior can help determine if the integral will approach a finite limit or not.

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