Question Number 59175 by maxmathsup by imad last updated on 05/May/19
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}\: \\ $$
Commented by maxmathsup by imad last updated on 07/May/19
![let I =∫_0 ^∞ ((1−cos(x))/x^2 ) dx ⇒I =∫_0 ^∞ ((2sin^2 ((x/2)))/x^2 ) dx =_((x/2)=t) 2 ∫_0 ^∞ ((sin^2 (t))/(4t^2 )) (2)dt =∫_0 ^∞ ((sin^2 t)/t^2 ) dt by parts u^′ =(1/t^2 ) and v =sin^2 t ⇒ I =[−(1/t)sin^2 (t)]_0 ^(+∞) −∫_0 ^∞ −(1/t) (2sint cost)dt =∫_0 ^∞ ((sin(2t))/t) dt =_(2t=u) ∫_0 ^∞ ((sin(u))/(u/2)) (du/2) =∫_0 ^∞ ((sinu)/u) du =(π/2) ⇒ ∫_0 ^∞ ((1−cosx)/x^2 ) dx =(π/2) .](https://www.tinkutara.com/question/Q59296.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({x}\right)}{{x}^{\mathrm{2}} }\:{dx}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:{dx}\:=_{\frac{{x}}{\mathrm{2}}={t}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}^{\mathrm{2}} \left({t}\right)}{\mathrm{4}{t}^{\mathrm{2}} }\:\left(\mathrm{2}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} }\:{dt}\:\:\:\:{by}\:{parts}\:\:{u}^{'} =\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:{and}\:{v}\:={sin}^{\mathrm{2}} {t}\:\Rightarrow \\ $$$${I}\:=\left[−\frac{\mathrm{1}}{{t}}{sin}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{+\infty} \:−\int_{\mathrm{0}} ^{\infty} \:\:−\frac{\mathrm{1}}{{t}}\:\left(\mathrm{2}{sint}\:{cost}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}\:{dt}\:=_{\mathrm{2}{t}={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({u}\right)}{\frac{{u}}{\mathrm{2}}}\:\frac{{du}}{\mathrm{2}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinu}}{{u}}\:{du}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }\:{dx}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$