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Question Number 91771 by  M±th+et+s last updated on 03/May/20

$${\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(x\right)}{x^2}dx$$

Commented by abdomathmax last updated on 03/May/20

$$letI={\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(x\right)}{x^2}dxwehsvebyparts$$$$I={[-\frac{1}{x}{sin}^{3}x]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{x}3{sin}^{2}xcosxdx$$$$=3{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}xcosx}{x}dx$$$$=\frac{3}{2}{\int }_0^{\mathrm{\infty }}\frac{cosxsin\left(2x\right)}{x}dxwehave$$$$cosxsin\left(2x\right)=cosxcos(\frac{\pi }{2}-2x)$$$$=\frac{1}{2}\{cos(-x+\frac{\pi }{2})+cosx(3x-\frac{\pi }{2})\}$$$$=\frac{1}{2}\{sinx+sin\left(3x\right)\}\Rightarrow$$$$I=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{sinx+sin\left(3x\right)}{x}dx$$$$=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx+\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{sin\left(3x\right)}{x}dxbut$$$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx=\frac{\pi }{2}$$$${\int }_0^{\mathrm{\infty }}\frac{sin\left(3x\right)}{x}dx{=}_{3x=t}{\int }_0^{\mathrm{\infty }}\frac{sint}{\frac{t}{3}}\times \frac{dt}{3}$$$$={\int }_0^{\mathrm{\infty }}\frac{sint}{t}dt=\frac{\pi }{2}\Rightarrow I=\frac{3}{4}\frac{\pi }{2}+\frac{3}{4}\frac{\pi }{2}$$$$=\frac{3\pi }{8}+\frac{3\pi }{8}=2\times \frac{3\pi }{8}=\frac{3\pi }{4}\Rightarrow I=\frac{3\pi }{4}$$

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