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Question Number 91534 by  M±th+et+s last updated on 01/May/20

$${\int }_1^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{x^2}dx$$

Commented by mathmax by abdo last updated on 01/May/20

$$wehave{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx={\int }_0^1\frac{{sin}^{2}x}{x^2}dx+{\int }_1^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx\Rightarrow$$$${\int }_1^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx={\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx-{\int }_0^1\frac{{sin}^{2}x}{x^2}dxwehavebyparts$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx={[-\frac{1}{x}{sin}^{2}x]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{x}\left(2sinx\right)cosxdx$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)}{x}dx{=}_{2x=t}{\int }_0^{\mathrm{\infty }}\frac{sint}{\frac{t}{2}}\times \frac{dt}{2}={\int }_0^{\mathrm{\infty }}\frac{sint}{t}dt=\frac{\pi }{2}$$$${\int }_0^1\frac{{sin}^{2}x}{x^2}dx={[-\frac{1}{x}{sin}^{2}x]}_0^1+{\int }_0^1\frac{1}{x}\left(2sinx\right)cosxdx$$$$=-{sin}^2\left(1\right)+{\int }_0^1\frac{sin\left(2x\right)}{x}dx=-{sin}^2\left(1\right)+{\int }_0^2\frac{sint}{t}dt(t=2x)$$$$wehavesint=t-\frac{t^3}{6}+o\left(t^3\right)\Rightarrow t-\frac{t^3}{6}⩽sint⩽t\Rightarrow$$$$1-\frac{t^2}{6}⩽\frac{sint}{t}⩽1\Rightarrow {\int }_0^2(1-\frac{t^2}{6})dt⩽{\int }_0^2\frac{sint}{t}dt⩽2\Rightarrow$$$${[t-\frac{t^3}{18}]}_0^2⩽{\int }_0^2\frac{sint}{t}dt⩽2\Rightarrow \frac{14}{9}⩽{\int }_0^2\frac{sint}{t}⩽2\Rightarrow$$$$v_0=\frac{7}{9}+1=\frac{16}{9}isapproximatevalueforthisintegral\Rightarrow$$$${\int }_1^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx\sim \frac{\pi }{2}+{sin}^2\left(1\right)-\frac{16}{9}$$$$$$

Commented by  M±th+et+s last updated on 01/May/20

$$greatsolutionthanxsir$$

Commented by mathmax by abdo last updated on 01/May/20

$$youarewelcome$$

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