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Question Number 129545 by mnjuly1970 last updated on 16/Jan/21

Answered by mindispower last updated on 16/Jan/21

$$={\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{x^2}-{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{1+x^2}dx$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{1+x^2}dx={\int }_0^{\mathrm{\infty }}\frac{1-cos\left(2x\right)}{2(1+x^2)}dx=\frac{\pi }{4}-\frac{1}{2}Re{\int }_0^{\mathrm{\infty }}\frac{e^{2ix}}{1+x^2}$$$$=\frac{\pi }{4}-\frac{Re}{2}.2i\pi .\frac{e^{-2}}{2i}=\frac{\pi }{4}-\frac{\pi }{2e^2}$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{x^2}={[-\frac{{sin}^2\left(x\right)}{x}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)}{x}dx$$$$=\frac{\pi }{2}$$$$weget\frac{\pi }{4}(1+\frac{2}{e^2})$$

Commented by mnjuly1970 last updated on 16/Jan/21

$$godkeepyou..mrpower..$$

Answered by mathmax by abdo last updated on 16/Jan/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{x}}^4}\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}={\int }_0^{\mathrm{\infty }}(\frac{1}{{\mathrm{x}}^2}-\frac{1}{{\mathrm{x}}^2+1}){\sin }^2\mathrm{xdx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}\mathrm{by}\mathrm{parts}$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2}\mathrm{dx}={[-\frac{{\sin }^2\mathrm{x}}{\mathrm{x}}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{x}}2\mathrm{sinx}\mathrm{cosxdx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\sin \left(2\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{\sin \left(2\mathrm{x}\right)}{2\mathrm{x}}\mathrm{d}\left(2\mathrm{x}\right)=\frac{\pi }{2}$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{1-\cos \left(2\mathrm{x}\right)}{2({\mathrm{x}}^2+1)}\mathrm{dx}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=\frac{\pi }{4}-\frac{1}{4}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}\mathrm{and}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{ix}}}{{\mathrm{x}}^2+1}\mathrm{dx}\right)$$$$=\mathrm{Re}\left(2\mathrm{i}\pi \frac{{\mathrm{e}}^{-2}}{2\mathrm{i}}\right)=\frac{\pi }{{\mathrm{e}}^2}\Rightarrow \mathrm{\Phi }=\frac{\pi }{2}-(\frac{\pi }{4}-\frac{\pi }{{4\mathrm{e}}^2})$$$$=\frac{\pi }{4}+\frac{\pi }{{4\mathrm{e}}^2}\Rightarrow \mathrm{\Phi }=\frac{\pi }{4}(1+\frac{1}{{\mathrm{e}}^2}).$$

Commented by mnjuly1970 last updated on 16/Jan/21

$$thanksalotmrmax...$$

Commented by mathmax by abdo last updated on 16/Jan/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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