Integration Questions

Question Number 124608 by TANMAY PANACEA last updated on 04/Dec/20

$$\int_{\mathrm{0}} ^{\infty} {sinx}^{{p}} \:{dx} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}^{{p}} }{{x}^{{q}} }{dx} \\$$ $${collected}\:{question} \\$$

Answered by Dwaipayan Shikari last updated on 04/Dec/20

$$\int_{\mathrm{0}} ^{\infty} {sinx}^{{p}} {dx} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {e}^{{ix}^{{p}} } −{e}^{−{ix}^{{p}} } {dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{p}} ={u} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}{ip}}\int_{\mathrm{0}} ^{\infty} \left({u}\right)^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{{iu}} −\frac{\mathrm{1}}{\mathrm{2}{ip}}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{−{iu}} {dt}\:\:\:\:\:\:\:\:\:\:{iu}=−\Psi\:\:\:\:\:{iu}=\varphi \\$$ $$=−\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{−\Psi}{{i}}\right)^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{−\Psi} {d}\Psi\:+\:\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\varphi}{{i}}\right)^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{−\varphi} {d}\varphi \\$$ $$=\frac{\left({i}\right)^{\frac{\mathrm{1}}{{p}}−\mathrm{1}} }{\mathrm{2}{p}}\Gamma\left(\frac{\mathrm{1}}{{p}}\right)+\frac{\left(−{i}\right)^{\frac{\mathrm{1}}{{p}}−\mathrm{1}} }{\mathrm{2}{p}}\Gamma\left(\frac{\mathrm{1}}{{p}}\right) \\$$ $$\frac{\mathrm{1}}{\mathrm{2}{p}}\Gamma\left(\frac{\mathrm{1}}{{p}}\right)\left(\frac{\mathrm{1}}{{i}}\left({e}^{\frac{\pi}{\mathrm{2}{p}}{i}} −{e}^{−\frac{\pi}{\mathrm{2}{p}}{i}} \right)\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{{p}}\right){sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)}{{p}}=\Gamma\left(\frac{\mathrm{1}}{{p}}+\mathrm{1}\right){sin}\left(\frac{\pi}{\mathrm{2}{p}}\right) \\$$

Commented byTANMAY PANACEA last updated on 04/Dec/20

$${excellent} \\$$

Answered by mathmax by abdo last updated on 04/Dec/20

$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{p}} \right)}{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{ix}^{\mathrm{p}} } }{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\right)\:\mathrm{but} \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{ix}^{\mathrm{p}} } }{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\:=_{\mathrm{ix}^{\mathrm{p}} =\mathrm{t}} \left(−\mathrm{i}\right)^{\mathrm{p}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{e}^{−\mathrm{t}} }{\left(\left(−\mathrm{it}\right)^{\frac{\mathrm{1}}{\mathrm{p}}} \right)^{\mathrm{q}} }\:\frac{\mathrm{1}}{\mathrm{p}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{p}}−\mathrm{1}} \:\mathrm{dt} \\$$ $$=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{e}^{−\frac{\mathrm{ip}\pi}{\mathrm{2}}} \:×\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{p}}−\mathrm{1}} }{\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{q}}{\mathrm{p}}} \:\mathrm{t}^{\frac{\mathrm{q}}{\mathrm{p}}} }\mathrm{e}^{−\mathrm{t}} \mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{e}^{−\frac{\mathrm{ip}\pi}{\mathrm{2}}} \:\mathrm{e}^{\frac{\mathrm{iq}\pi}{\mathrm{2p}}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{p}}−\frac{\mathrm{q}}{\mathrm{p}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\$$ $$=\frac{\mathrm{1}}{\mathrm{p}}\:\mathrm{e}^{\mathrm{i}\left\{\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right\}} \:\:\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right)\:\:\:\left(\mathrm{so}\:\mathrm{o}<\mathrm{q}<\mathrm{1}\right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{p}}\left\{\mathrm{cos}\left(\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right)\right\}\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right)\:\Rightarrow \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{p}} \right)}{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{p}}\mathrm{sin}\left(\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{sin}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}−\frac{\mathrm{q}\pi}{\mathrm{2p}}\right)×\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right) \\$$