Question Number 32994 by abdo imad last updated on 09/Apr/18
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left(\lambda{x}\right)}{{x}^{\mathrm{2}} }\:{dx}\:{with}\:\lambda>\mathrm{0}\:. \\ $$
Commented byabdo imad last updated on 15/Apr/18
![changement λx =u give I = ∫_0 ^∞ ((1−cosu)/(((u/λ))^2 )) (du/λ) =λ ∫_0 ^∞ ((1−cosu)/u^2 ) du but ∫_0 ^∞ ((1−cosu)/u^2 ) du = 2 ∫_0 ^∞ ((sin^2 ((u/2)))/u^2 )du =_((u/2)=t) 2 ∫_0 ^∞ ((sin^2 t)/(4t^2 )) 2dt =∫_0 ^∞ ((sin^2 t)/t^2 )dt (by parts) = [−(1/t) sin^2 t]_0 ^(+∞) −∫_0 ^∞ −(1/t) 2sint cost dt = ∫_0 ^∞ ((sin(2t))/t) dt = _(2t=x) ∫_0 ^∞ ((sin(x))/(x/2)) (dx/2) = ∫_0 ^(+∞) ((sinx)/x)dx =(π/2) ⇒ I = ((λπ)/2) .](Q33404.png)
$${changement}\:\:\lambda{x}\:={u}\:{give} \\ $$ $${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cosu}}{\left(\frac{{u}}{\lambda}\right)^{\mathrm{2}} }\:\frac{{du}}{\lambda}\:\:=\lambda\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cosu}}{{u}^{\mathrm{2}} }\:{du}\:{but} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cosu}}{{u}^{\mathrm{2}} }\:{du}\:\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{sin}^{\mathrm{2}} \left(\frac{{u}}{\mathrm{2}}\right)}{{u}^{\mathrm{2}} }{du} \\ $$ $$=_{\frac{{u}}{\mathrm{2}}={t}} \:\:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}^{\mathrm{2}} {t}}{\mathrm{4}{t}^{\mathrm{2}} }\:\mathrm{2}{dt}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} }{dt}\:\:\:\left({by}\:{parts}\right) \\ $$ $$=\:\left[−\frac{\mathrm{1}}{{t}}\:{sin}^{\mathrm{2}} {t}\right]_{\mathrm{0}} ^{+\infty} \:\:\:−\int_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{{t}}\:\mathrm{2}{sint}\:{cost}\:{dt} \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}\:{dt}\:=\:_{\mathrm{2}{t}={x}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({x}\right)}{\frac{{x}}{\mathrm{2}}}\:\frac{{dx}}{\mathrm{2}}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{sinx}}{{x}}{dx} \\ $$ $$=\frac{\pi}{\mathrm{2}}\:\:\Rightarrow\:\:{I}\:\:=\:\frac{\lambda\pi}{\mathrm{2}}\:. \\ $$