Question Number 91534 by M±th+et+s last updated on 01/May/20
$$\int_{\mathrm{1}} ^{\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 01/May/20
![we have ∫_0 ^∞ ((sin^2 x)/x^2 )dx =∫_0 ^1 ((sin^2 x)/x^2 )dx +∫_1 ^∞ ((sin^2 x)/x^2 )dx ⇒ ∫_1 ^∞ ((sin^2 x)/x^2 )dx =∫_0 ^∞ ((sin^2 x)/x^2 )dx−∫_0 ^1 ((sin^2 x)/x^2 )dx we have by parts ∫_0 ^∞ ((sin^2 x)/x^2 )dx =[−(1/x)sin^2 x]_0 ^∞ +∫_0 ^∞ (1/x)(2sinx)cosx dx = ∫_0 ^∞ ((sin(2x))/x)dx =_(2x=t) ∫_0 ^∞ ((sint)/(t/2))×(dt/2) =∫_0 ^∞ ((sint)/t)dt =(π/2) ∫_0 ^1 ((sin^2 x)/x^2 )dx =[−(1/x)sin^2 x]_0 ^1 +∫_0 ^1 (1/x)(2sinx)cosx dx =−sin^2 (1)+∫_0 ^1 ((sin(2x))/x)dx =−sin^2 (1)+ ∫_0 ^2 ((sint)/t)dt (t=2x) we have sint =t−(t^3 /6) +o(t^3 ) ⇒t−(t^3 /6)≤sint ≤t ⇒ 1−(t^2 /6)≤((sint)/t)≤1 ⇒ ∫_0 ^2 (1−(t^2 /6))dt≤∫_0 ^2 ((sint)/t)dt≤2 ⇒ [t−(t^3 /(18))]_0 ^2 ≤ ∫_0 ^2 ((sint)/t)dt ≤2 ⇒((14)/9) ≤ ∫_0 ^2 ((sint)/t) ≤2 ⇒ v_0 =(7/9) +1 =((16)/9) is approximate value for this integral ⇒ ∫_1 ^∞ ((sin^2 x)/x^2 )dx ∼ (π/2) +sin^2 (1)−((16)/9)](Q91589.png)
$${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:+\int_{\mathrm{1}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:\:{we}\:{have}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:=\left[−\frac{\mathrm{1}}{{x}}{sin}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{x}}\left(\mathrm{2}{sinx}\right){cosx}\:{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:=_{\mathrm{2}{x}={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sint}}{\frac{{t}}{\mathrm{2}}}×\frac{{dt}}{\mathrm{2}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sint}}{{t}}{dt}\:=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:=\left[−\frac{\mathrm{1}}{{x}}{sin}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{x}}\left(\mathrm{2}{sinx}\right){cosx}\:{dx} \\ $$$$=−{sin}^{\mathrm{2}} \left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:=−{sin}^{\mathrm{2}} \left(\mathrm{1}\right)+\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{sint}}{{t}}{dt}\:\:\:\:\:\left({t}=\mathrm{2}{x}\right) \\ $$$${we}\:{have}\:{sint}\:={t}−\frac{{t}^{\mathrm{3}} }{\mathrm{6}}\:+{o}\left({t}^{\mathrm{3}} \right)\:\Rightarrow{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{6}}\leqslant{sint}\:\leqslant{t}\:\Rightarrow \\ $$$$\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{6}}\leqslant\frac{{sint}}{{t}}\leqslant\mathrm{1}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{6}}\right){dt}\leqslant\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{sint}}{{t}}{dt}\leqslant\mathrm{2}\:\Rightarrow \\ $$$$\left[{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{18}}\right]_{\mathrm{0}} ^{\mathrm{2}} \:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{sint}}{{t}}{dt}\:\leqslant\mathrm{2}\:\Rightarrow\frac{\mathrm{14}}{\mathrm{9}}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{sint}}{{t}}\:\leqslant\mathrm{2}\:\:\Rightarrow \\ $$$${v}_{\mathrm{0}} =\frac{\mathrm{7}}{\mathrm{9}}\:+\mathrm{1}\:=\frac{\mathrm{16}}{\mathrm{9}}\:{is}\:{approximate}\:{value}\:{for}\:{this}\:{integral}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:\sim\:\:\frac{\pi}{\mathrm{2}}\:+{sin}^{\mathrm{2}} \left(\mathrm{1}\right)−\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$ \\ $$
Commented by M±th+et+s last updated on 01/May/20
$${great}\:{solution}\:{thanx}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 01/May/20
$${you}\:{are}\:{welcome} \\ $$