$$\left.{a}\right)\:\:{let}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}} \:{x}^{−\alpha} \:{dx}\:\:\Rightarrow{f}\left(\alpha\right)\:=_{{ix}\:={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{{t}} \:\left(\frac{{t}}{{i}}\right)^{−\alpha} \:\frac{{dt}}{{i}} \\$$$$=\frac{\mathrm{1}}{{i}^{\mathrm{1}−\alpha} }\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{{t}} \:{t}^{\mathrm{1}−\alpha−\mathrm{1}} \:{dt}\:=\frac{\mathrm{1}}{{i}^{\mathrm{1}−\alpha} }\Gamma\left(\mathrm{1}−\alpha\right)\:{the}\:{convergence}\:{of}\:{f}\left(\alpha\right) \\$$$${is}\:{assured} \\$$$$\left.{b}\right)\int_{\mathrm{0}} ^{\infty} \:{cosx}\:{x}^{−\alpha} \:{dx}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}} \:{x}^{−\alpha} {dx}\right)\:={Re}\left(\frac{\mathrm{1}}{{i}^{\alpha−\mathrm{1}} }\Gamma\left(\mathrm{1}−\alpha\right)\right)\:{but} \\$$$$\frac{\mathrm{1}}{{i}^{\alpha−\mathrm{1}} }\Gamma\left(\mathrm{1}−\alpha\right)\:=\frac{\mathrm{1}}{\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{\alpha−\mathrm{1}} }\Gamma\left(\mathrm{1}−\alpha\right)\:={e}^{−\frac{{i}\pi}{\mathrm{2}}\left(\alpha−\mathrm{1}\right)} \Gamma\left(\mathrm{1}−\alpha\right) \\$$$$={e}^{\frac{{i}\pi}{\mathrm{2}}} \:{e}^{−\frac{{i}\pi}{\mathrm{2}}\alpha} \:\Gamma\left(\mathrm{1}−\alpha\right)\:={i}\left({cos}\left(\frac{\pi\alpha}{\mathrm{2}}\right)−{isin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\right)\Gamma\left(\mathrm{1}−\alpha\right) \\$$$$={sin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\alpha\right)+{i}\:{cos}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\alpha\right)\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} {cosx}\:{x}^{−\alpha} {dx}\:={sin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\alpha\right) \\$$$$\\$$
$${thank}\:{you}\:{sir}.{but}\:{the}\:{first}\:{question} \\$$$$\:{ineed}\:{to}\:{prove}\:{the}\:{integral}\:{is}\:{ineed} \\$$$${convergent}\:{before}\:{computing}\:{its}\:{value} \\$$$${concerning}\:.{the}\:{second}\:{question},{i}\:{meant}\:{by}\:{complex}\:{integration} \\$$$${is}\:{by}\:{using}\:{the}\:{residue}\:{theorem} \\$$$$\\$$