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Question Number 94114 by naka3546 last updated on 16/May/20

$$Find\underset{1}{\int }\stackrel{\mathrm{\infty }}{}\frac{{\sin }^{2}x}{x^2}dx$$

Commented by mathmax by abdo last updated on 17/May/20

$$I={\int }_1^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dxwehave{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx={\int }_0^1\frac{{sin}^{2}x}{x^2}dx+{\int }_1^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx$$$$\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx-{\int }_0^1\frac{{sin}^{2}x}{x^2}dxbut$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx{=}_{byparts}{[-\frac{1}{x}{sin}^{2}x]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{x}\left(2sinxcosx\right)dx$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{x}sin\left(2x\right)dx{=}_{2x=t}{\int }_0^{\mathrm{\infty }}\frac{sint}{\frac{t}{2}}\times \frac{dt}{2}={\int }_0^{\mathrm{\infty }}\frac{sint}{t}dt=\frac{\pi }{2}\Rightarrow$$$$I=\frac{\pi }{2}-{\int }_0^1\frac{1-cos\left(2x\right)}{x^2}dxwehave$$$$cos\left(2x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}{\left(2x\right)}^{2n}=\sum _{n=0}^{\mathrm{\infty }}\frac{2^{2n}{(-1)}^n}{\left(2n\right)!}{x}^{2n}$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n{4}^n}{\left(2n\right)!}{x}^{2n}\Rightarrow 1-cosx=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n{4}^n}{\left(2n\right)!}{x}^{2n}\Rightarrow$$$$\frac{1-cosx}{x^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{4}^n}{\left(2n\right)!}{x}^{2n-2}\Rightarrow$$$${\int }_0^1\frac{1-cosx}{x^2}dx=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{4}^n}{\left(2n\right)!}{\left[\frac{1}{2n-1}{x}^{2n-1}\right]}_0^1$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{4}^n}{(2n-1)\left(2n\right)!}\Rightarrow I=\frac{\pi }{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n{4}^n}{(2n-1)\left(2n\right)!}$$$$I=\frac{\pi }{2}-\frac{4}{2!}+\frac{4^2}{3\left(4\right)!}-\frac{4^3}{5\left(6\right)!}+....$$$$wecantaken=4or5forapproximatevalue.$$

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