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Question Number 159931 by qaz last updated on 22/Nov/21

$$\mathrm{Prove}::{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{\mathrm{n}}\mathrm{x}}{{\mathrm{x}}^{\mathrm{m}}}\mathrm{dx}=\frac{1}{\mathrm{\Gamma }\left(\mathrm{m}\right)}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{D}}^{\mathrm{m}-1}{\sin }^{\mathrm{n}}\mathrm{x}}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{n}+\mathrm{m}\in \mathrm{Odd}\mathrm{Number}.$$

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