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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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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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