Question Number 34315 by prof Abdo imad last updated on 03/May/18
$$\left.\mathrm{1}\right)\:{find}\:\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{e}^{−{at}} \:−{e}^{−{bt}} }{{t}}{sin}\left({xt}\right){dt} \\ $$$${with}\:{a}>\mathrm{0}\:,{b}>\mathrm{0}\:. \\ $$
Commented by math khazana by abdo last updated on 05/May/18
![we have F^′ (x) = ∫_0 ^∞ (e^(−at) −e^(−bt) ) cos(xt)dt =Re( ∫_0 ^∞ (e^(−at) −e^(−bt) )e^(ixt) dt) =Re( ∫_0 ^∞ (e^((−a +ix)t) −e^((−b +ix)t) )dt but ∫_0 ^∞ e^((−a+ix)t) dt = [ (1/(−a +ix)) e^((−1+ix)t) ]_0 ^(+∞) = (1/(−a +ix)) =((−1)/(a−ix)) =−((a+ix)/(a^2 +x^2 )) by the same manner ∫_0 ^∞ e^((−b+ix)t) dt = −((b +ix)/(b^2 +x^2 )) ⇒ F^′ (x) = Re( ((b+ix)/(b^2 +x^2 )) −((a+ix)/(a^2 +x^2 ))) =(b/(b^2 +x^2 )) − (a/(a^2 +x^2 )) ⇒ F(x)= ∫_0 ^x (b/(b^2 +t^2 ))dt −∫_0 ^x (a/(a^2 +t^2 ))dt +λ but ∫_0 ^x (b/(b^2 +t^2 ))dt =_(t=bu) ∫_0 ^(x/b) (b/(b^2 (1+u^2 ))) bdu = arctan((x/b)) ⇒F(x)= arctan((x/b)) −arctan((x/a)) +λ λ =F(0) ⇒F(x)=arctan((x/b)) −arctan((x/a)) .](https://www.tinkutara.com/question/Q34415.png)
$${we}\:{have}\:{F}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\left({e}^{−{at}} \:−{e}^{−{bt}} \right)\:{cos}\left({xt}\right){dt} \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:\left({e}^{−{at}} \:−{e}^{−{bt}} \right){e}^{{ixt}} {dt}\right) \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:\left({e}^{\left(−{a}\:+{ix}\right){t}} \:−{e}^{\left(−{b}\:+{ix}\right){t}} \right){dt}\:\:{but}\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{\left(−{a}+{ix}\right){t}} {dt}\:=\:\:\left[\:\frac{\mathrm{1}}{−{a}\:+{ix}}\:{e}^{\left(−\mathrm{1}+{ix}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\:\frac{\mathrm{1}}{−{a}\:+{ix}}\:=\frac{−\mathrm{1}}{{a}−{ix}}\:=−\frac{{a}+{ix}}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:\:{by}\:{the}\:{same}\:{manner} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−{b}+{ix}\right){t}} {dt}\:\:=\:−\frac{{b}\:+{ix}}{{b}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}^{'} \left({x}\right)\:=\:{Re}\left(\:\frac{{b}+{ix}}{{b}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:−\frac{{a}+{ix}}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\right) \\ $$$$=\frac{{b}}{{b}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:−\:\frac{{a}}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:\Rightarrow\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\frac{{b}}{{b}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dt} \\ $$$$−\int_{\mathrm{0}} ^{{x}} \:\:\frac{{a}}{{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dt}\:+\lambda\:\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{b}}{{b}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dt}\:=_{{t}={bu}} \:\:\int_{\mathrm{0}} ^{\frac{{x}}{{b}}} \:\:\:\frac{{b}}{{b}^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:{bdu} \\ $$$$=\:{arctan}\left(\frac{{x}}{{b}}\right)\:\Rightarrow{F}\left({x}\right)=\:{arctan}\left(\frac{{x}}{{b}}\right)\:−{arctan}\left(\frac{{x}}{{a}}\right)\:+\lambda \\ $$$$\lambda\:={F}\left(\mathrm{0}\right)\:\Rightarrow{F}\left({x}\right)={arctan}\left(\frac{{x}}{{b}}\right)\:−{arctan}\left(\frac{{x}}{{a}}\right)\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 08/May/18
$${we}\:{know}\:{that}\:{arctan}\:\alpha\:−{arctan}\beta\:={arctan}\left(\frac{\alpha−\beta}{\mathrm{1}+\alpha\beta}\right) \\ $$$$\Rightarrow{F}\left({x}\right)\:=\:{arctan}\left(\frac{\frac{{x}}{{b}}\:−\frac{{x}}{{a}}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{ab}}}\right) \\ $$$$=\:{arctan}\left(\frac{{ax}−{bx}}{{x}^{\mathrm{2}} \:\:+{ab}}\:\right)\:. \\ $$