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Question Number 65729 by aliesam last updated on 02/Aug/19

Commented by mathmax by abdo last updated on 03/Aug/19

$$a)letf\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{e}^{ix}{x}^{-\alpha }dx\Rightarrow f\left(\alpha \right){=}_{ix=t}{\int }_0^{\mathrm{\infty }}{e}^t{\left(\frac{t}{i}\right)}^{-\alpha }\frac{dt}{i}$$$$=\frac{1}{i^{1-\alpha }}{\int }_0^{\mathrm{\infty }}{e}^t{t}^{1-\alpha -1}dt=\frac{1}{i^{1-\alpha }}\mathrm{\Gamma }(1-\alpha )theconvergenceoff\left(\alpha \right)$$$$isassured$$$$b){\int }_0^{\mathrm{\infty }}cosx{x}^{-\alpha }dx=Re\left({\int }_0^{\mathrm{\infty }}{e}^{ix}{x}^{-\alpha }dx\right)=Re\left(\frac{1}{i^{\alpha -1}}\mathrm{\Gamma }(1-\alpha )\right)but$$$$\frac{1}{i^{\alpha -1}}\mathrm{\Gamma }(1-\alpha )=\frac{1}{{\left(e^{i\frac{\pi }{2}}\right)}^{\alpha -1}}\mathrm{\Gamma }(1-\alpha )=e^{-\frac{i\pi }{2}(\alpha -1)}\mathrm{\Gamma }(1-\alpha )$$$$=e^{\frac{i\pi }{2}}{e}^{-\frac{i\pi }{2}\alpha }\mathrm{\Gamma }(1-\alpha )=i(cos\left(\frac{\pi \alpha }{2}\right)-isin\left(\frac{\pi \alpha }{2}\right))\mathrm{\Gamma }(1-\alpha )$$$$=sin\left(\frac{\pi \alpha }{2}\right)\mathrm{\Gamma }(1-\alpha )+icos\left(\frac{\pi \alpha }{2}\right)\mathrm{\Gamma }(1-\alpha )\Rightarrow$$$${\int }_0^{\mathrm{\infty }}cosx{x}^{-\alpha }dx=sin\left(\frac{\pi \alpha }{2}\right)\mathrm{\Gamma }(1-\alpha )$$$$$$

Commented by aliesam last updated on 03/Aug/19

$$thankyousir.butthefirstquestion$$$$ineedtoprovetheintegralisineed$$$$convergentbeforecomputingitsvalue$$$$concerning.thesecondquestion,imeantbycomplexintegration$$$$isbyusingtheresiduetheorem$$$$$$

Commented by aliesam last updated on 04/Aug/19

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