Question Number 117380 by mnjuly1970 last updated on 11/Oct/20
$$\:\:\:\:\:\:\:\:\:\:\:...\:\:{prove}\:\:{that}\:... \\ $$$$\:\: \\ $$$$\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{sin}\left(\pi^{\mathrm{2}} {x}+\frac{\mathrm{1}}{{x}}\right){dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}\pi}} \\ $$$$ \\ $$$$\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$
Commented by mindispower last updated on 11/Oct/20
$${nice}\:{one} \\ $$
Answered by mnjuly1970 last updated on 13/Oct/20
![solution:: Recall ::∫_0 ^( ∞) sin(z^2 )dz =^(..fresnel integral..) (√(π/8)) Ω =^(t=(√x)) ∫_0 ^( ∞) sin(π^2 t^2 +(1/t^2 ))dt πΩ=∫_0 ^( ∞) πsin(π^2 t^2 +(1/t^2 ))dt (i) πΩ =^(t=(1/(πu))) (1/π) ∫_(0 ) ^( ∞) πsin((1/u^2 )+π^2 u^2 )(du/u^2 ) πΩ=∫_0 ^( ∞) sin(π^2 u^2 +(1/u^2 ))(du/u^2 ) (ii) (i)+(ii) :: 2πΩ =∫_0 ^( ∞) (π+(1/x^2 ))sin[(πx−(1/x))^2 +2π]du 2πΩ=^(πx−(1/x)=y) ∫_(−∞) ^( ∞) sin(y^2 )dy 2πΩ =^(Recall ) 2 (√(π/8)) ⇒ Ω =(√(1/(8π))) ✓ ...♣M.N.july.1970♣... ♠peace be upon you♠](Q117725.png)
$${solution}:: \\ $$$$\mathscr{R}{ecall}\:::\int_{\mathrm{0}} ^{\:\infty} {sin}\left({z}^{\mathrm{2}} \right){dz}\:\overset{..{fresnel}\:\:{integral}..} {=}\sqrt{\frac{\pi}{\mathrm{8}}} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\Omega\:\overset{{t}=\sqrt{{x}}} {=}\:\int_{\mathrm{0}} ^{\:\infty} {sin}\left(\pi^{\mathrm{2}} {t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}\:\: \\ $$$$\:\:\:\pi\Omega=\int_{\mathrm{0}} ^{\:\infty} \pi{sin}\left(\pi^{\mathrm{2}} {t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}\:\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\pi\Omega\:\:\:\overset{{t}=\frac{\mathrm{1}}{\pi{u}}} {=}\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}\:} ^{\:\infty} \pi{sin}\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\pi^{\mathrm{2}} {u}^{\mathrm{2}} \right)\frac{{du}}{{u}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\pi\Omega=\int_{\mathrm{0}} ^{\:\infty} {sin}\left(\pi^{\mathrm{2}} {u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right)\frac{{du}}{{u}^{\mathrm{2}} }\:\:\:\left({ii}\right) \\ $$$$\:\:\:\:\left({i}\right)+\left({ii}\right)\:::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\pi\Omega\:=\int_{\mathrm{0}} ^{\:\infty} \left(\pi+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){sin}\left[\left(\pi{x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\pi\right]{du} \\ $$$$\:\:\:\:\mathrm{2}\pi\Omega\overset{\pi{x}−\frac{\mathrm{1}}{{x}}={y}} {=}\int_{−\infty} ^{\:\:\infty} {sin}\left({y}^{\mathrm{2}} \right){dy} \\ $$$$\:\:\:\:\mathrm{2}\pi\Omega\:\overset{\mathscr{R}{ecall}\:} {=}\mathrm{2}\:\sqrt{\frac{\pi}{\mathrm{8}}}\:\:\Rightarrow\:\Omega\:=\sqrt{\frac{\mathrm{1}}{\mathrm{8}\pi}}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\clubsuit\mathscr{M}.\mathscr{N}.{july}.\mathrm{1970}\clubsuit... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\spadesuit{peace}\:\:{be}\:\:{upon}\:\:{you}\spadesuit \\ $$$$\:\: \\ $$