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Question Number 163033 by mnjuly1970 last updated on 03/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x-sin\left(x\right)}{x^3}dx$$$$---solution---$$$$\mathrm{\Omega }\stackrel{\mathcal{I}.\mathcal{B}.\mathcal{P}}{=}{\left[\frac{-1}{2x^2}(x-sin\left(x\right))\right]}_0^{\mathrm{\infty }}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(x\right)}{x^2}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{2{sin}^2\left(\frac{x}{2}\right)}{x^2}dx={\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(\frac{x}{2}\right)}{x^2}dx$$$$\stackrel{\frac{x}{2}=\alpha }{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(\alpha \right)}{{\alpha }^2}d\alpha$$$$=\frac{1}{2}{\left[\frac{-1}{\alpha }{sin}^2\left(\alpha \right)\right]}_0^{\mathrm{\infty }}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2\alpha \right)}{\alpha }d\alpha$$$$\stackrel{2\alpha =\phi }{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(\phi \right)}{\phi }d\phi =\frac{\pi }{4}$$$$--\mathrm{\Omega }=\frac{\pi }{4}---$$$$$$$$$$

Answered by qaz last updated on 03/Jan/22

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\mathcal{L}\{\mathrm{x}-\sin \mathrm{x}\}\cdot {\mathcal{L}}^{-1}\left\{\frac{1}{{\mathrm{x}}^3}\right\}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}(\frac{1}{{\mathrm{x}}^2}-\frac{1}{1+{\mathrm{x}}^2})\cdot \frac{{\mathrm{x}}^2}{2!}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{\pi }{4}$$

Commented by mnjuly1970 last updated on 03/Jan/22

$$niceverynice..laplacetransform$$

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