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Question Number 155237 by mnjuly1970 last updated on 27/Sep/21

$$$$$$\mathrm{I}f\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(x\right).sin\left(\sqrt{x}\right)}{x}dx$$$$provethat:$$$$\mathrm{\Omega }=4{\gamma }^2+\frac{{\pi }^3}{3}◼m.n$$

Answered by phanphuoc last updated on 27/Sep/21

$$puty=\sqrt{x}\to \mathrm{\Omega }=8{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}ysiny}{y}dy$$$$\mathrm{\Omega }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{x}^{-\alpha }sinxdx$$$$mellintransformM\left(sinx\right)=\mathrm{\Gamma }\left(\alpha \right)sin\frac{\pi \alpha }{2}=\mathrm{\Omega }\left(\alpha \right)$$$$\to \mathrm{\Omega }=d^2\left(\mathrm{\Omega }\left(\alpha \right)\right)/d{\alpha }^2=......$$

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