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Question Number 143312 by Ar Brandon last updated on 12/Jun/21

$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^4\mathrm{x}}{{\mathrm{x}}^4}\mathrm{dx}=\frac{\pi }{3}$$

Answered by Olaf_Thorendsen last updated on 12/Jun/21

$$\mathrm{Let}f\left(x\right)=1\left(\mathrm{constant}\mathrm{function}\mathrm{unity}\right)$$$$\mathrm{As}f\mathrm{is}\mathrm{a}\mathrm{continuous}\mathrm{function}\mathrm{satisfying}$$$$\mathrm{the}\pi -\mathrm{periodic}\mathrm{assumption},\mathrm{we}\mathrm{can}$$$$\mathrm{apply}\mathrm{the}\mathrm{extended}{\mathrm{Lobachevsky}}^{\prime }\mathrm{s}$$$$\mathrm{Dirichlet}\mathrm{integral}\mathrm{formula}:$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^4}{x^4}f\left(x\right)dx={\int }_0^{\frac{\pi }{2}}f\left(x\right)dx-\frac{2}{3}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x.f\left(x\right)dx$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{{\sin }^4}{x^4}dx={\int }_0^{\frac{\pi }{2}}dx-\frac{2}{3}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^4}{x^4}dx=\frac{\pi }{2}-\frac{2}{3}{\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dx$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^4}{x^4}dx=\frac{\pi }{2}-\frac{1}{3}{[x-\frac{1}{2}\sin 2x]}_0^{\frac{\pi }{2}}$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^4}{x^4}dx=\frac{\pi }{2}-\frac{1}{3}\left(\frac{\pi }{2}\right)$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^4}{x^4}dx=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$$

Commented by Ar Brandon last updated on 12/Jun/21

$$\mathrm{Thanks}\mathrm{Sir}$$

Commented by Ar Brandon last updated on 12/Jun/21

$$\mathrm{I}\mathrm{tried}\mathrm{this};$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\sin }^4\mathrm{x}}{{\mathrm{x}}^4}\mathrm{dx},$$$${\sin }^4\mathrm{x}=\frac{(1-2\cos 2\mathrm{x}+{\cos }^{2}2\mathrm{x})}{4}=\frac{1}{4}-\frac{\cos 2\mathrm{x}}{2}+\frac{(1+\cos 4\mathrm{x})}{8}$$$$\mathrm{\Phi }=\frac{3}{8}{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{x}}^4}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\cos 2\mathrm{x}}{{\mathrm{x}}^4}\mathrm{dx}+\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{\cos 4\mathrm{x}}{{\mathrm{x}}^4}\mathrm{dx}$$$$=-{\left[\frac{1}{{8\mathrm{x}}^3}\right]}_0^{\mathrm{\infty }}-\frac{\pi 2^3}{4\mathrm{\Gamma }\left(4\right)\cos \left(\frac{\pi }{2}\times 4\right)}+\frac{\pi 4^3}{16\mathrm{\Gamma }\left(4\right)\cos \left(\frac{\pi }{2}\times 4\right)}$$$$=-{\left[\frac{1}{{8\mathrm{x}}^3}\right]}_0^{\mathrm{\infty }}-\frac{8\pi }{24}+\frac{64\pi }{48}=...$$$$\mathrm{Why}\mathrm{did}\mathrm{it}\mathrm{not}\mathrm{go}?\mathrm{Do}\mathrm{you}\mathrm{know}\mathrm{why}?$$

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