Question Number 34314 by prof Abdo imad last updated on 03/May/18
$${let}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} {dt}\: \\ $$$${calculate}\:{f}\left({x}\right)\:. \\ $$
Commented by prof Abdo imad last updated on 06/May/18
![we have f^′ (x) = ∫_0 ^∞ ((t sin(xt))/t^2 ) e^(−t) dt = ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt ⇒ f^(′′) (x)= ∫_0 ^∞ ((t cos(xt))/t) e^(−t) dt =∫_0 ^∞ e^(−t) cos(xt)dt =Re( ∫_0 ^∞ e^(−t) e^(ixt) dt) =Re( ∫_0 ^∞ e^((−1+ix)t) dt) but ∫_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 )) ⇒ f^(′′) (x)= (1/(1+x^2 )) ⇒ f^′ (x)= arctanx +λ but λ =f^′ (0)=0 ⇒ f^′ (x)= arctanx ⇒ f(x) = ∫ arctanxdx +c ∫ arctanxdx = x arctanx −∫ (x/(1+x^2 ))dx = x arctanx −(1/2)ln(1+x^2 ) ⇒ f(x)= x arctanx −(1/2)ln(1+x^2 ) +c c =f(0) =0 ⇒ f(x)= x arctanx −(1/2) ln(1+x^2 ) .](https://www.tinkutara.com/question/Q34443.png)
$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}\:{sin}\left({xt}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} \:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({xt}\right)}{{t}}\:{e}^{−{t}} \:{dt}\:\Rightarrow\:{f}^{''} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}\:{cos}\left({xt}\right)}{{t}}\:{e}^{−{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} \:{cos}\left({xt}\right){dt}\:={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:{e}^{{ixt}} {dt}\right) \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−\mathrm{1}+{ix}\right){t}} {dt}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−\mathrm{1}+{ix}\right){t}} {dt}\:=\left[\:\frac{\mathrm{1}}{−\mathrm{1}\:+{ix}}{e}^{\left(−\mathrm{1}+{ix}\right){t}} \right]_{\mathrm{0}} ^{+\infty} =\frac{−\mathrm{1}}{−\mathrm{1}+{ix}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−{ix}}\:=\:\frac{\mathrm{1}+{ix}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow\:{f}^{''} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:{arctanx}\:+\lambda\:{but}\:\lambda\:={f}^{'} \left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:{arctanx}\:\Rightarrow\:{f}\left({x}\right)\:=\:\int\:{arctanxdx}\:+{c} \\ $$$$\int\:\:{arctanxdx}\:=\:{x}\:{arctanx}\:−\int\:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\: \\ $$$$=\:{x}\:{arctanx}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\:{x}\:{arctanx}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+{c} \\ $$$${c}\:={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow\: \\ $$$${f}\left({x}\right)=\:{x}\:{arctanx}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:. \\ $$