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Question Number 34315 by prof Abdo imad last updated on 03/May/18

$$1)findF\left(x\right)={\int }_0^{+\mathrm{\infty }}\frac{e^{-at}-e^{-bt}}{t}sin\left(xt\right)dt$$ $$witha>0,b>0.$$

Commented bymath khazana by abdo last updated on 05/May/18

$$wehave{F}^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}(e^{-at}-e^{-bt})cos\left(xt\right)dt$$ $$=Re\left({\int }_0^{\mathrm{\infty }}(e^{-at}-e^{-bt})e^{ixt}dt\right)$$ $$=Re({\int }_0^{\mathrm{\infty }}(e^{(-a+ix)t}-e^{(-b+ix)t})dtbut$$ $${\int }_0^{\mathrm{\infty }}{e}^{(-a+ix)t}dt={\left[\frac{1}{-a+ix}{e}^{(-1+ix)t}\right]}_0^{+\mathrm{\infty }}$$ $$=\frac{1}{-a+ix}=\frac{-1}{a-ix}=-\frac{a+ix}{a^2+x^2}bythesamemanner$$ $${\int }_0^{\mathrm{\infty }}{e}^{(-b+ix)t}dt=-\frac{b+ix}{b^2+x^2}\Rightarrow$$ $$F^{{}^{\prime }}\left(x\right)=Re(\frac{b+ix}{b^2+x^2}-\frac{a+ix}{a^2+x^2})$$ $$=\frac{b}{b^2+x^2}-\frac{a}{a^2+x^2}\Rightarrow F\left(x\right)={\int }_0^x\frac{b}{b^2+t^2}dt$$ $$-{\int }_0^x\frac{a}{a^2+t^2}dt+\lambda but$$ $${\int }_0^x\frac{b}{b^2+t^2}dt{=}_{t=bu}{\int }_0^{\frac{x}{b}}\frac{b}{b^2(1+u^2)}bdu$$ $$=arctan\left(\frac{x}{b}\right)\Rightarrow F\left(x\right)=arctan\left(\frac{x}{b}\right)-arctan\left(\frac{x}{a}\right)+\lambda$$ $$\lambda =F\left(0\right)\Rightarrow F\left(x\right)=arctan\left(\frac{x}{b}\right)-arctan\left(\frac{x}{a}\right).$$ $$$$

Commented bymath khazana by abdo last updated on 08/May/18

$$weknowthatarctan\alpha -arctan\beta =arctan\left(\frac{\alpha -\beta }{1+\alpha \beta }\right)$$ $$\Rightarrow F\left(x\right)=arctan\left(\frac{\frac{x}{b}-\frac{x}{a}}{1+\frac{x^2}{ab}}\right)$$ $$=arctan\left(\frac{ax-bx}{x^2+ab}\right).$$

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