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Question Number 152653 by mnjuly1970 last updated on 30/Aug/21

$$$$$$\mathrm{\Omega }:={\int }_0^{\mathrm{\infty }}\frac{e^{-x^3}.sin\left(x^3\right)}{x}dx=\frac{\zeta \left(2\right)}{2}$$$$m.n...$$$$$$

Answered by qaz last updated on 30/Aug/21

$$\mathrm{\Omega }=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}\sin \mathrm{x}}{\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{3}{\int }_0^1\mathrm{ds}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{sx}}\sin \mathrm{xdx}+\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{\sin \mathrm{x}}{\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{3}{\int }_0^1\frac{1}{{\mathrm{s}}^2+1}\mathrm{ds}+\frac{\pi }{6}$$$$=-\frac{1}{3}\cdot \frac{\pi }{4}+\frac{\pi }{6}$$$$=\frac{\pi }{12}$$

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