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Question Number 190672 by 073 last updated on 08/Apr/23

Answered by aba last updated on 08/Apr/23

$$\frac{1}{4}(1+\frac{1}{{\mathrm{e}}^2})\pi ?$$

Commented by 073 last updated on 08/Apr/23

$$\mathrm{solution}\mathrm{please}??$$

Answered by qaz last updated on 09/Apr/23

$${\int }_0^{\mathrm{\infty }}\frac{\sin {}^{2}x}{x^2(x^2+1)}dx=\frac{\pi }{2}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1-\cos 2x}{x^2+1}dx$$$$=\frac{\pi }{4}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\cos 2x}{x^2+1}dx$$$$=\frac{\pi }{4}+\frac{1}{2}{\mathcal{L}}^{-1}{\left\{{\int }_0^{\mathrm{\infty }}\frac{\mathcal{L}\left\{\cos tx\right)}{x^2+1}dx\right\}}_{t=2}$$$$=\frac{\pi }{4}+\frac{1}{2}{\mathcal{L}}^{-1}{\left\{{\int }_0^{\mathrm{\infty }}\frac{s}{(s^2+x^2)(x^2+1)}dx\right\}}_{t=2}$$$$=\frac{\pi }{4}+\frac{\pi }{4}{\mathcal{L}}^{-1}{\left\{\frac{1}{s+1}\right\}}_{t=2}=\frac{\pi }{4}{(1+e^{-t})}_{t=2}=\frac{\pi }{4}(1+e^{-2})$$

Commented by aba last updated on 09/Apr/23

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