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Question Number 134147 by Ñï= last updated on 28/Feb/21

$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^{1}xsin\left(2\pi nx\right)dx=?$$

Answered by mathmax by abdo last updated on 28/Feb/21

$$\mathrm{let}{\mathrm{u}}_{\mathrm{n}}={\int }_0^1\mathrm{xsin}\left(2\pi \mathrm{nx}\right)\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{for}\mathrm{n}>0$$$${\mathrm{u}}_{\mathrm{n}}={[-\frac{\mathrm{x}}{2\pi \mathrm{n}}\cos \left(2\pi \mathrm{nx}\right)]}_0^1+{\int }_0^1\frac{1}{2\pi \mathrm{n}}\cos \left(2\pi \mathrm{nx}\right)\mathrm{dx}$$$$=\frac{1}{4{\pi }^2{\mathrm{n}}^2}{\left[\sin \left(2\pi \mathrm{nx}\right)\right]}_0^1=0\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=0$$

Answered by mnjuly1970 last updated on 28/Feb/21

$$method1$$$$2\pi nx=t$$$$\mathrm{\Omega }={lim}_{n\to \mathrm{\infty }}\frac{1}{4{\pi }^2{n}^n}{\int }_0^{2\pi n}tsin\left(t\right)dt$$$$={lim}_{n\to \mathrm{\infty }}\frac{1}{4{\pi }^2{n}^2}{[-tcost+sin\left(t\right)\left(t\right)]}_0^{2\pi n}$$$$={lim}_{n\to \mathrm{\infty }}\frac{-1}{2\pi n}=0$$$$method2$$$$f:[a,b]\to \mathbb{R}is\ast Reimanintegrable\ast then:$$$${lim}_{t\to \mathrm{\infty }}{\int }_a^{b}f\left(x\right)sin\left(xt\right)dx=0$$$$reiman-lebesguetheorem...$$$$$$

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