# Show that $\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \sin (x^2) \, dx = \frac{(-1)^k}{\xi_k}$ for a suitable $\xi_k$

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.

• $\xi_k$ appears on the right side of the equation but not the left; do you mean 'for some $\xi_k$ with $\sqrt{k\pi}\lt\xi_k\lt\sqrt{(k+1)\pi}$'? Feb 8, 2013 at 1:02
• 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. Feb 8, 2013 at 1:11
• @StevenStadnicki Yes, I suppose I meant to say that. @ julien Yes, I'm currently trying to understand Ivan's solution. Feb 8, 2013 at 1:41

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 $y=x^2$, $$\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \sin{(x^2)} \, dx =\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{y}} \, dy$$
$$\frac{1}{\sqrt{(k+1)\pi}}=\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{(k+1)\pi}} \, dy \right|<\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{y}} \, dy \right|<\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{k\pi}} \, dy \right|=\frac{1}{\sqrt{k\pi}}$$
• Note that $\left|\int_{k\pi}^{(k+1)\pi} sin{y} \, dy \right|=\left|(-cos{(k+1)\pi})-(-cos{k\pi}) \right|=2$, thus $\left|\int_{k\pi}^{(k+1)\pi} \frac{\sin{y}}{2 \sqrt{k\pi}} \, dy \right|=\frac{\left|\int_{k\pi}^{(k+1)\pi} sin{y} \, dy \right|}{2 \sqrt{k\pi}}=\frac{1}{\sqrt{k\pi}}$ Feb 8, 2013 at 1:43