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Question Number 161945 by Ahmed777hamouda last updated on 24/Dec/21

$$\mathit{P}\mathbf{rove}\mathbf{that}{\int }_0^{\mathrm{\infty }}\frac{{(\mathit{x}\mathit{c}\mathit{o}\mathit{s}\left(\mathit{x}\right)-\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right))}^2}{{\mathit{x}}^6}\mathit{d}\mathit{x}=\frac{\mathit{\pi }}{15}$$

Answered by Ar Brandon last updated on 24/Dec/21

$$I={\int }_0^{\mathrm{\infty }}\frac{{(x\cos x-\sin x)}^2}{x^6}dx={\int }_0^{\mathrm{\infty }}(\frac{{\cos }^{2}x}{x^4}-\frac{\sin 2x}{x^5}+\frac{{\sin }^{2}x}{x^6})dx$$$$={[-\frac{{\cos }^{2}x}{3x^3}]}_0^{\mathrm{\infty }}-\frac{1}{3}\int \frac{\sin 2x}{x^3}dx-16{\int }_0^{\mathrm{\infty }}\frac{\sin u}{u^5}du-{\left[\frac{{\sin }^{2}x}{5x^5}\right]}_0^{\mathrm{\infty }}+\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{\sin 2x}{x^5}dx$$$$=-\frac{4}{3}{\int }_0^{\mathrm{\infty }}\frac{\sin t}{t^3}dt-16{\int }_0^{\mathrm{\infty }}\frac{\sin u}{u^5}du+\frac{16}{5}{\int }_0^{\mathrm{\infty }}\frac{\sin t}{t^5}dt$$$$=-\frac{4}{3}\cdot \frac{\pi }{2\mathrm{\Gamma }\left(3\right)\sin \left(\frac{3\pi }{2}\right)}-\frac{64}{5}\cdot \frac{\pi }{2\mathrm{\Gamma }\left(5\right)\sin \left(\frac{5\pi }{2}\right)}=\frac{\pi }{3}-\frac{4\pi }{15}=\frac{\pi }{15}$$

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