Question Number 26563 by abdo imad last updated on 26/Dec/17
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }\:{dx} \\ $$
Commented by abdo imad last updated on 28/Dec/17
![let put I= ∫_0 ^∞ ((1−cosx )/x^2 )dx=lim_(ε−>0^+ ) ∫_ε ^∝ ((1−cosx)/x^2 )dx= =lim_(ε−>0^+ ) I(ε) we integr by parties I(ε)= [ −(1/x)(1−cosx)]_ε ^∝ − ∫_ε ^∝ −(1/x)sinxdx = (1/ε)(1−cosε) + ∫_ε ^∝ ((sinx)/x)dx but lim_(ε−>0) (1/ε)(1−cosε)=lim_(ε−>0) 2ε(((sinε)/ε))^2 =0 and lim_(ε−>0) ∫_ε ^∝ ((sinx)/x)dx= (π/2) ⇒ I= (π/2).](https://www.tinkutara.com/question/Q26676.png)
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cosx}\:\:}{{x}^{\mathrm{2}} }{dx}={lim}_{\varepsilon−>\mathrm{0}^{+} } \:\int_{\varepsilon} ^{\propto} \:\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }{dx}= \\ $$$$={lim}_{\varepsilon−>\mathrm{0}^{+} } \:\:{I}\left(\varepsilon\right)\:{we}\:{integr}\:{by}\:{parties} \\ $$$${I}\left(\varepsilon\right)=\:\left[\:−\frac{\mathrm{1}}{{x}}\left(\mathrm{1}−{cosx}\right)\right]_{\varepsilon} ^{\propto} \:\:−\:\int_{\varepsilon} ^{\propto} −\frac{\mathrm{1}}{{x}}{sinxdx} \\ $$$$=\:\frac{\mathrm{1}}{\varepsilon}\left(\mathrm{1}−{cos}\varepsilon\right)\:+\:\int_{\varepsilon} ^{\propto} \frac{{sinx}}{{x}}{dx}\:\:{but} \\ $$$${lim}_{\varepsilon−>\mathrm{0}} \:\frac{\mathrm{1}}{\varepsilon}\left(\mathrm{1}−{cos}\varepsilon\right)={lim}_{\varepsilon−>\mathrm{0}} \mathrm{2}\varepsilon\left(\frac{{sin}\varepsilon}{\varepsilon}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${and}\:{lim}_{\varepsilon−>\mathrm{0}} \:\int_{\varepsilon} ^{\propto} \frac{{sinx}}{{x}}{dx}=\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}=\:\frac{\pi}{\mathrm{2}}. \\ $$
Answered by prakash jain last updated on 27/Dec/17
$$−\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}}\right)+\int\frac{\mathrm{sin}\:{x}}{{x}}{dx} \\ $$$$−\frac{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}}+\mathrm{Si}\left({x}\right)+{C} \\ $$$${taking}\:{limits} \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$