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Question Number 128110 by Dwaipayan Shikari last updated on 04/Jan/21

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{sinxsin(x+y)}{x(x+y)}dxdy$$

Answered by Olaf last updated on 04/Jan/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\sin x\sin (x+y)}{x(x+y)}dxdy$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}\left({\int }_0^{\mathrm{\infty }}\frac{\sin (x+y)}{x+y}dy\right)dx$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}\left({\int }_x^{\mathrm{\infty }}\frac{\sin u}{u}du\right)dx$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}({\int }_0^{\mathrm{\infty }}\frac{\sin u}{u}du-{\int }_0^x\frac{\sin u}{u}du)dx$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}(\frac{\pi }{2}-\mathrm{Si}\left(x\right))dx$$$$\mathrm{\Omega }=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx-{\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}\mathrm{Si}\left(x\right)dx$$$$\mathrm{\Omega }=\frac{{\pi }^2}{4}-{\int }_0^{\mathrm{\infty }}{\mathrm{Si}}^{\prime }\left(x\right)\mathrm{Si}\left(x\right)dx$$$$\mathrm{\Omega }=\frac{{\pi }^2}{4}-\frac{1}{2}{\left[{\mathrm{Si}}^2\left(x\right)\right]}_0^{\mathrm{\infty }}$$$$\mathrm{\Omega }=\frac{{\pi }^2}{4}-\frac{1}{2}{\left(\underset{x\to \mathrm{\infty }}{\lim }\mathrm{Si}\left(x\right)\right)}^2$$$$\mathrm{\Omega }=\frac{{\pi }^2}{4}-\frac{1}{2}.\frac{{\pi }^2}{4}$$$$\mathrm{\Omega }=\frac{{\pi }^2}{8}$$

Commented by Dwaipayan Shikari last updated on 04/Jan/21

$$Greatsir!$$

Commented by BHOOPENDRA last updated on 05/Jan/21

$$greatsir$$

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