Question Number 35630 by abdo mathsup 649 cc last updated on 21/May/18
$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} {dt} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left({t}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} \:{dt}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
![by parts u^′ = (1/t^2 ) and v =(1−cos(xt))e^(−t) ⇒ v^′ = xsin(xt) e^(−t) −(1−cos(xt))e^(−t) =( x sin(xt) +cos(xt)−1) e^(−t) f(x) = [ −(1/t) (1−cos(xt))e^(−t) ]_0 ^(+∞) −∫_0 ^∞ ((−1)/t) ( xsin(xt) +cos(xt) −1)e^(−t) dt but 1−cos(xt) ∼ ((x^2 t^2 )/2) ⇒ ((1−cos(xt))/t) ∼ ((tx^2 )/2) ⇒ lim_(t→0) (((1−cos(xt))e^(−t) )/t) =0 ⇒ f(x) = x ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt +∫_0 ^∞ ((1−cos(xt))/t) e^(−t) dt let calculate h(x)= ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt h^′ (x) = ∫_0 ^∞ cos(xt) e^(−t) dt =Re(∫_0 ^∞ e^(ixt−t) dt) ∫_0 ^∞ e^((−1 +ix)t) dt = [ (1/(−1+ix)) e^((−1+ix)t) ]_0 ^(+∞) = (1/(−1+ix)) = ((−1)/(1−ix)) = ((−1−ix)/(1+x^2 )) ⇒ h^′ (x) = ((−1)/(1+x^2 )) ⇒ h(x) =λ −arctan(x) but λ =h(0)=0 ⇒ h(x) =−arctan(x)](Q35813.png)
$${by}\:{parts}\:\:{u}^{'} \:=\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\:{and}\:{v}\:=\left(\mathrm{1}−{cos}\left({xt}\right)\right){e}^{−{t}} \:\:\Rightarrow \\ $$$${v}^{'} \:=\:{xsin}\left({xt}\right)\:{e}^{−{t}} \:−\left(\mathrm{1}−{cos}\left({xt}\right)\right){e}^{−{t}} \\ $$$$=\left(\:{x}\:{sin}\left({xt}\right)\:+{cos}\left({xt}\right)−\mathrm{1}\right)\:{e}^{−{t}} \\ $$$${f}\left({x}\right)\:=\:\left[\:−\frac{\mathrm{1}}{{t}}\:\left(\mathrm{1}−{cos}\left({xt}\right)\right){e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{−\mathrm{1}}{{t}}\:\left(\:{xsin}\left({xt}\right)\:+{cos}\left({xt}\right)\:−\mathrm{1}\right){e}^{−{t}} \:{dt}\:{but} \\ $$$$\mathrm{1}−{cos}\left({xt}\right)\:\sim\:\frac{{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}}\:\:\sim\:\:\frac{{tx}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} \:\frac{\left(\mathrm{1}−{cos}\left({xt}\right)\right){e}^{−{t}} }{{t}}\:=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\:{x}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({xt}\right)}{{t}}\:{e}^{−{t}} \:{dt}\:\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}}\:{e}^{−{t}} {dt} \\ $$$${let}\:{calculate}\:{h}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{sin}\left({xt}\right)}{{t}}\:{e}^{−{t}} \:{dt}\: \\ $$$${h}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{cos}\left({xt}\right)\:{e}^{−{t}} \:{dt}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{{ixt}−{t}} {dt}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{\left(−\mathrm{1}\:+{ix}\right){t}} {dt}\:=\:\left[\:\frac{\mathrm{1}}{−\mathrm{1}+{ix}}\:{e}^{\left(−\mathrm{1}+\boldsymbol{{ix}}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\:\frac{\mathrm{1}}{−\mathrm{1}+{ix}}\:=\:\frac{−\mathrm{1}}{\mathrm{1}−{ix}}\:=\:\frac{−\mathrm{1}−{ix}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow\:{h}^{'} \left({x}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{h}\left({x}\right)\:=\lambda\:−{arctan}\left({x}\right)\:\:{but}\:\lambda\:={h}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${h}\left({x}\right)\:=−{arctan}\left({x}\right) \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
$${let}\:{calculate}\:{K}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}\:−{cos}\left({xt}\right)}{{t}}\:{e}^{−{t}} {dt} \\ $$$${K}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{sin}\left({xt}\right)\:{e}^{−{t}} {dt}\:={Im}\left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{{ixt}\:−{t}} {dt}\right) \\ $$$$=\:{Im}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−\mathrm{1}+{ix}\right){t}} {dt}\right)\:=\:\frac{−{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${K}\left({x}\right)=\:\lambda\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\:{but}\:\:\lambda\:={K}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:−{x}\:{arctan}\left({x}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
$${error}\:{in}\:{the}\:{final}\:{line}\: \\ $$$${f}\left({x}\right)=\:−{x}\:{arctan}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right). \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} \:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:−{x}\:{arctan}\left({x}\right) \\ $$$${let}\:{take}\:{x}=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−{cos}\left({t}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} =\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:−\frac{\pi}{\mathrm{4}}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
$${error}\:{in}\:{line}\:\mathrm{8} \\ $$$${f}\left({x}\right)=\:{x}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({xt}\right)}{{t}}\:{e}^{−{t}} \:{dt}\:−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}}{e}^{−{t}} \:{dt} \\ $$