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Question Number 124608 by TANMAY PANACEA last updated on 04/Dec/20
∫_0 ^∞ sinx^p dx ∫_0 ^∞ ((sinx^p )/x^q )dx collected question
$$\int_{\mathrm{0}} ^{\infty} {sinx}^{{p}} \:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}^{{p}} }{{x}^{{q}} }{dx} \\ $$$${collected}\:{question} \\ $$
Answered by Dwaipayan Shikari last updated on 04/Dec/20
∫_0 ^∞ sinx^p dx =(1/(2i))∫_0 ^∞ e^(ix^p ) −e^(−ix^p ) dx x^p =u =(1/(2ip))∫_0 ^∞ (u)^((1−p)/p) e^(iu) −(1/(2ip))∫_0 ^∞ u^((1−p)/p) e^(−iu) dt iu=−Ψ iu=ϕ =−(1/(2p))∫_0 ^∞ (((−Ψ)/i))^((1−p)/p) e^(−Ψ) dΨ + (1/(2p))∫_0 ^∞ ((ϕ/i))^((1−p)/p) e^(−ϕ) dϕ =(((i)^((1/p)−1) )/(2p))Γ((1/p))+(((−i)^((1/p)−1) )/(2p))Γ((1/p)) (1/(2p))Γ((1/p))((1/i)(e^((π/(2p))i) −e^(−(π/(2p))i) ))=((Γ((1/p))sin((π/(2p))))/p)=Γ((1/p)+1)sin((π/(2p)))
$$\int_{\mathrm{0}} ^{\infty} {sinx}^{{p}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {e}^{{ix}^{{p}} } −{e}^{−{ix}^{{p}} } {dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{p}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ip}}\int_{\mathrm{0}} ^{\infty} \left({u}\right)^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{{iu}} −\frac{\mathrm{1}}{\mathrm{2}{ip}}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{−{iu}} {dt}\:\:\:\:\:\:\:\:\:\:{iu}=−\Psi\:\:\:\:\:{iu}=\varphi \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{−\Psi}{{i}}\right)^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{−\Psi} {d}\Psi\:+\:\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\varphi}{{i}}\right)^{\frac{\mathrm{1}−{p}}{{p}}} {e}^{−\varphi} {d}\varphi \\ $$$$=\frac{\left({i}\right)^{\frac{\mathrm{1}}{{p}}−\mathrm{1}} }{\mathrm{2}{p}}\Gamma\left(\frac{\mathrm{1}}{{p}}\right)+\frac{\left(−{i}\right)^{\frac{\mathrm{1}}{{p}}−\mathrm{1}} }{\mathrm{2}{p}}\Gamma\left(\frac{\mathrm{1}}{{p}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{p}}\Gamma\left(\frac{\mathrm{1}}{{p}}\right)\left(\frac{\mathrm{1}}{{i}}\left({e}^{\frac{\pi}{\mathrm{2}{p}}{i}} −{e}^{−\frac{\pi}{\mathrm{2}{p}}{i}} \right)\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{{p}}\right){sin}\left(\frac{\pi}{\mathrm{2}{p}}\right)}{{p}}=\Gamma\left(\frac{\mathrm{1}}{{p}}+\mathrm{1}\right){sin}\left(\frac{\pi}{\mathrm{2}{p}}\right) \\ $$
Commented by TANMAY PANACEA last updated on 04/Dec/20
excellent
$${excellent} \\ $$
Answered by mathmax by abdo last updated on 04/Dec/20
∫_0 ^∞ ((sin(x^p ))/x^q )dx =−Im(∫_0 ^∞ (e^(−ix^p ) /x^q )dx) but ∫_0 ^∞ (e^(−ix^p ) /x^q )dx =_(ix^p =t) (−i)^p ∫_0 ^∞ (e^(−t) /(((−it)^(1/p) )^q )) (1/p)t^((1/p)−1) dt =(1/p)e^(−((ipπ)/2)) ×∫_0 ^∞ (t^((1/p)−1) /((e^(−((iπ)/2)) )^(q/p) t^(q/p) ))e^(−t) dt =(1/p)e^(−((ipπ)/2)) e^((iqπ)/(2p)) ∫_0 ^∞ t^((1/p)−(q/p)−1) e^(−t) dt =(1/p) e^(i{((qπ)/(2p))−((pπ)/2)}) Γ(((1−q)/p)) (so o<q<1) =(1/p){cos(((qπ)/(2p))−((pπ)/2))+isin(((qπ)/(2p))−((pπ)/2))}Γ(((1−q)/p)) ⇒ ∫_0 ^∞ ((sin(x^p ))/x^q )dx =−(1/p)sin(((qπ)/(2p))−((pπ)/2))Γ(((1−q)/p)) =(1/p)sin(((pπ)/2)−((qπ)/(2p)))×Γ(((1−q)/p))
$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{p}} \right)}{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{ix}^{\mathrm{p}} } }{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\right)\:\mathrm{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{ix}^{\mathrm{p}} } }{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\:=_{\mathrm{ix}^{\mathrm{p}} =\mathrm{t}} \left(−\mathrm{i}\right)^{\mathrm{p}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{e}^{−\mathrm{t}} }{\left(\left(−\mathrm{it}\right)^{\frac{\mathrm{1}}{\mathrm{p}}} \right)^{\mathrm{q}} }\:\frac{\mathrm{1}}{\mathrm{p}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{p}}−\mathrm{1}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{e}^{−\frac{\mathrm{ip}\pi}{\mathrm{2}}} \:×\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{p}}−\mathrm{1}} }{\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{q}}{\mathrm{p}}} \:\mathrm{t}^{\frac{\mathrm{q}}{\mathrm{p}}} }\mathrm{e}^{−\mathrm{t}} \mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{e}^{−\frac{\mathrm{ip}\pi}{\mathrm{2}}} \:\mathrm{e}^{\frac{\mathrm{iq}\pi}{\mathrm{2p}}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{p}}−\frac{\mathrm{q}}{\mathrm{p}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{p}}\:\mathrm{e}^{\mathrm{i}\left\{\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right\}} \:\:\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right)\:\:\:\left(\mathrm{so}\:\mathrm{o}<\mathrm{q}<\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{p}}\left\{\mathrm{cos}\left(\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right)\right\}\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{p}} \right)}{\mathrm{x}^{\mathrm{q}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{p}}\mathrm{sin}\left(\frac{\mathrm{q}\pi}{\mathrm{2p}}−\frac{\mathrm{p}\pi}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{sin}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}−\frac{\mathrm{q}\pi}{\mathrm{2p}}\right)×\Gamma\left(\frac{\mathrm{1}−\mathrm{q}}{\mathrm{p}}\right) \\ $$

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