Question Number 167230 by mnjuly1970 last updated on 10/Mar/22
![lim_( α→∞) { (α ∫_0 ^( ∞) sin( x^( α) ) dx )=ϕ(α)]= (π/2) −−−− ∫_0 ^( ∞) sin(x^( α) )dx =^(x^( α) = y) (1/α)∫_0 ^( ∞) (( sin(y))/y^( 1−(1/α)) ) dy ⇒ α ∫_0 ^( ∞) sin(x^( α) ) dx = ∫_0 ^( ∞) (( sin(y))/y^( 1−(1/α)) ) dy = (( π)/(2 Γ (1−(1/α))sin ((π/2) (1−(1/α))))) = (π/(2Γ (1−(1/α))cos ((π/(2α)) ))) = ϕ (α ) lim_( α→∞) ϕ (α )=^((1/α) =β) lim_( β→0) (π/(2Γ (1−β)cos ((π/2) β))) = (π/2)](https://www.tinkutara.com/question/Q167230.png)
$$ \\ $$$$\:\:\:\:\:{lim}_{\:\alpha\rightarrow\infty} \left\{\:\left(\alpha\:\int_{\mathrm{0}} ^{\:\infty} {sin}\left(\:{x}^{\:\alpha} \right)\:{dx}\:\right)=\varphi\left(\alpha\right)\right]=\:\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:−−−− \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} {sin}\left({x}^{\:\alpha} \right){dx}\:\overset{{x}^{\:\alpha} =\:{y}} {=}\:\frac{\mathrm{1}}{\alpha}\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}\left({y}\right)}{{y}^{\:\mathrm{1}−\frac{\mathrm{1}}{\alpha}} }\:{dy} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\alpha\:\int_{\mathrm{0}} ^{\:\infty} {sin}\left({x}^{\:\alpha} \right)\:{dx}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}\left({y}\right)}{{y}^{\:\mathrm{1}−\frac{\mathrm{1}}{\alpha}} }\:{dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\:\pi}{\mathrm{2}\:\Gamma\:\left(\mathrm{1}−\frac{\mathrm{1}}{\alpha}\right){sin}\:\left(\frac{\pi}{\mathrm{2}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\alpha}\right)\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}\Gamma\:\left(\mathrm{1}−\frac{\mathrm{1}}{\alpha}\right){cos}\:\left(\frac{\pi}{\mathrm{2}\alpha}\:\right)}\:=\:\varphi\:\left(\alpha\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{lim}_{\:\alpha\rightarrow\infty} \:\varphi\:\left(\alpha\:\right)\overset{\frac{\mathrm{1}}{\alpha}\:=\beta} {=}{lim}_{\:\beta\rightarrow\mathrm{0}} \:\:\frac{\pi}{\mathrm{2}\Gamma\:\left(\mathrm{1}−\beta\right){cos}\:\left(\frac{\pi}{\mathrm{2}}\:\beta\right)}\:=\:\frac{\pi}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$