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Let $k \in \mathbb{N}$. Show that there exists $\sqrt{k\pi} < \xi_k < \sqrt{(k+1)\pi}$ such that:

$$\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \sin(x^2) \, dx = \frac{(-1)^k}{\xi_k}$$

We haven't introduced Fresnel integrals, which might be useful here... I guess I can't use them, though.

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    $\begingroup$ $\xi_k$ appears on the right side of the equation but not the left; do you mean 'for some $\xi_k$ with $\sqrt{k\pi}\lt\xi_k\lt\sqrt{(k+1)\pi}$'? $\endgroup$ Feb 8, 2013 at 1:02
  • $\begingroup$ Since $\sin(y)$ is of sign $(-1)^k$ on $(k\pi,(k+1)\pi)$, the sign of your integral is indeed $(-1)^k$. Now it remains to prove that $1/\sqrt{(k+1)\pi}<|\mbox{your integral}|<1/\sqrt{k\pi}$. I know, that's the meat of the question. $\endgroup$
    – Julien
    Feb 8, 2013 at 1:11
  • $\begingroup$ @StevenStadnicki Yes, I suppose I meant to say that. @ julien Yes, I'm currently trying to understand Ivan's solution. $\endgroup$
    – looki
    Feb 8, 2013 at 1:41

1 Answer 1

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Note that the integral is negative for odd $k$ and positive for even $k$, justifying the $(-1)^k$.

It thus suffices to show that $$\frac{1}{\sqrt{(k+1)\pi}}<\left|\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \sin{(x^2)} \, dx \right|<\frac{1}{\sqrt{k\pi}}$$

Do a change of variables $y=x^2$, $$\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \sin{(x^2)} \, dx =\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{y}} \, dy$$

$$\frac{1}{\sqrt{(k+1)\pi}}=\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{(k+1)\pi}} \, dy \right|<\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{y}} \, dy \right|<\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{k\pi}} \, dy \right|=\frac{1}{\sqrt{k\pi}}$$

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  • $\begingroup$ Nicely done, +1. $\endgroup$
    – Julien
    Feb 8, 2013 at 1:31
  • $\begingroup$ Thanks for your reply, but I have a problem (it's basically the core problem of this question). How do I prove the two equalities in the last step (fraction equals absolute value of integral). I have absolutely no idea what you did there. $\endgroup$
    – looki
    Feb 8, 2013 at 1:37
  • $\begingroup$ Note that $\left|\int_{k\pi}^{(k+1)\pi} sin{y} \, dy \right|=\left|(-cos{(k+1)\pi})-(-cos{k\pi}) \right|=2$, thus $\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{k\pi}} \, dy \right|=\frac{\left|\int_{k\pi}^{(k+1)\pi} sin{y} \, dy \right|}{2 \sqrt{k\pi}}=\frac{1}{\sqrt{k\pi}}$ $\endgroup$
    – Ivan Loh
    Feb 8, 2013 at 1:43
  • $\begingroup$ Ah, I get it now! Thanks a lot. I never would have been able to come up with this. $\endgroup$
    – looki
    Feb 8, 2013 at 1:54

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