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Question Number 142256 by rs4089 last updated on 28/May/21

$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x^{1-a}}dx$$

Answered by Dwaipayan Shikari last updated on 28/May/21

$${\int }_0^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x^{\mu }}dx=\frac{\pi }{2\mathrm{\Gamma }\left(\mu \right)sin\left(\frac{\mu }{2}\pi \right)}$$$${\int }_0^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x^{1-a}}=\frac{\pi }{2\mathrm{\Gamma }(1-a)cos\left(a\frac{\pi }{2}\right)}$$

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