Question Number 28989 by abdo imad last updated on 02/Feb/18
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }{dx}. \\ $$
Commented by abdo imad last updated on 04/Feb/18
![let put I= ∫_0 ^∞ ((sin^2 (3x))/x^2 )dx the ch .3x=t give I= ∫_0 ^∞ 9((sin^2 (t))/t^2 ) (1/3)dt = 3∫_0 ^∞ ((sin^2 t)/t^2 ) dt but ∫_0 ^∞ ((sin^2 t)/t^2 )dt=lim_(ξ→0) ∫_ξ ^(+∞) ((sin^2 t)/t^2 )dt by parts ∫_ξ ^(+∞) ((sin^2 t)/t^2 )dt= [((−1)/t)sin^2 t]_ξ ^(+∞) − ∫_ξ ^(+∞) ( −(1/t))2 sint costdt = ((sin^2 ξ)/ξ) + ∫_ξ ^(+∞) ((sin(2t))/t)dt→_(ξ→0) ∫_0 ^∞ ((sin(2t))/t)dt but ∫_0 ^∞ ((sin(2t))/t)dt= ∫_0 ^∞ 2 ((sinu)/u) (du/2) (ch. 2t=u) = (π/2) ⇒ I= ((3π)/2) .](https://www.tinkutara.com/question/Q29109.png)
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }{dx}\:\:{the}\:{ch}\:.\mathrm{3}{x}={t}\:{give} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\mathrm{9}\frac{{sin}^{\mathrm{2}} \left({t}\right)}{{t}^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{3}}{dt}\:=\:\mathrm{3}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} }\:{dt}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} }{dt}={lim}_{\xi\rightarrow\mathrm{0}} \:\int_{\xi} ^{+\infty} \:\:\frac{{sin}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} }{dt}\:\:{by}\:{parts} \\ $$$$\int_{\xi} ^{+\infty} \:\:\frac{{sin}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} }{dt}=\:\left[\frac{−\mathrm{1}}{{t}}{sin}^{\mathrm{2}} {t}\right]_{\xi} ^{+\infty} \:−\:\int_{\xi} ^{+\infty} \left(\:−\frac{\mathrm{1}}{{t}}\right)\mathrm{2}\:{sint}\:{costdt} \\ $$$$=\:\frac{{sin}^{\mathrm{2}} \xi}{\xi}\:\:+\:\int_{\xi} ^{+\infty} \:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}{dt}\rightarrow_{\xi\rightarrow\mathrm{0}} \int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}{dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}{dt}=\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{2}\:\frac{{sinu}}{{u}}\:\:\frac{{du}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({ch}.\:\mathrm{2}{t}={u}\right) \\ $$$$=\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\:\:{I}=\:\:\frac{\mathrm{3}\pi}{\mathrm{2}}\:. \\ $$