# Evaluate-0-sinx-x-a-dx-

Question Number 134474 by rs4089 last updated on 04/Mar/21
$${Evaluate}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{{a}} }{dx} \\$$
Answered by Dwaipayan Shikari last updated on 04/Mar/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{{a}} }{dx} \\$$$$=\frac{\mathrm{1}}{\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{xt}} {sinxdtdx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{x}\left({t}−{i}\right)} −{t}^{{a}−\mathrm{1}} {e}^{−{x}\left({t}+{i}\right)} {dx}\:{dt} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{{t}−{i}}−\frac{{t}^{{a}−\mathrm{1}} }{{t}+{i}}{dt} \\$$$$=\frac{\mathrm{1}}{\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{{a}}{\mathrm{2}}−\mathrm{1}} }{{u}+\mathrm{1}}{du}\:\:\:\:\:\:{t}^{\mathrm{2}} ={u} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left({a}\right)}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)} \\$$
Answered by mnjuly1970 last updated on 04/Mar/21
$$\:\:\:\boldsymbol{\phi}=−{im}\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{ix}} }{{x}^{{a}} }{dx}=−{im}\int_{\mathrm{0}\:} ^{\:\infty} {e}^{−{ix}} {x}^{−{a}} {dx} \\$$$$\:\:\:\:=−{im}\left(\mathscr{L}\:\left[{x}^{−{a}} \right]\right)\mid_{{s}={i}} \\$$$$\:\:\:\:\:=−{im}\left\{\frac{\Gamma\left(\mathrm{1}−{a}\right)}{{s}^{\mathrm{1}−{a}} }\mid_{{s}={i}} \right\} \\$$$$\:\:\:\:=−{im}\left\{\frac{\Gamma\left({a}\right)\Gamma\left(\mathrm{1}−{a}\right)}{\Gamma\left({a}\right){e}^{\frac{{i}\pi}{\mathrm{2}}\left(\mathrm{1}−{a}\right)} }\right\}\:\:\: \\$$$$\:\:\:\:\:\:\:=−{im}\left\{\frac{\pi}{\Gamma\left({a}\right){sin}\left(\pi{a}\right)\left({sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)+{icos}\left(\frac{\pi}{\mathrm{2}}{a}\right)\right)}\right. \\$$$$=−{im}\left\{\frac{\pi\left({sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)−{icos}\left(\frac{\pi}{\mathrm{2}}{a}\right)\right)}{\Gamma\left({a}\right){sin}\left(\pi{a}\right)}\right\} \\$$$$=\frac{\pi{cos}\left(\frac{\pi}{\mathrm{2}}{a}\right)}{\mathrm{2}\Gamma\left({a}\right){sin}\left(\frac{\pi}{\mathrm{2}}{a}\right){cos}\left(\frac{\pi}{\mathrm{2}}{a}\right)}=\frac{\pi}{\mathrm{2}\Gamma\left({a}\right){sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)}\:… \\$$$$\:\:\:\:\:\: \\$$
Answered by mathmax by abdo last updated on 27/Mar/21
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{a}} }\mathrm{dx}\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{ix}} \:\mathrm{dx}\right)\:\mathrm{we}\:\mathrm{have} \\$$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{ix}} \:\mathrm{dx}\:\:=_{\mathrm{ix}=\mathrm{t}\rightarrow\mathrm{x}=−\mathrm{it}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{it}\right)^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{t}} \:\left(−\mathrm{i}\right)\mathrm{dt} \\$$$$=\left(−\mathrm{i}\right)^{−\mathrm{a}+\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{1}−\mathrm{a}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\left(\mathrm{1}−\mathrm{a}\right)} \:\Gamma\left(\mathrm{1}−\mathrm{a}\right) \\$$$$=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}\left(\mathrm{a}−\mathrm{1}\right)} \:.\Gamma\left(\mathrm{1}−\mathrm{a}\right)\:=\Gamma\left(\mathrm{1}−\mathrm{a}\right)\:\left\{\mathrm{cos}\left(\frac{\pi\left(\mathrm{a}−\mathrm{1}\right)}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\pi\left(\mathrm{a}−\mathrm{1}\right)}{\mathrm{2}}\right)\right\}\:\Rightarrow \\$$$$\Phi=−\Gamma\left(\mathrm{1}−\mathrm{a}\right)\mathrm{sin}\left(\frac{\pi\left(\mathrm{a}−\mathrm{1}\right)}{\mathrm{2}}\right)\:=\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{a}\right)\right)×\Gamma\left(\mathrm{1}−\mathrm{a}\right) \\$$