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Question Number 29849 by abdo imad last updated on 12/Feb/18

$$letgivea>0,b>0findthevslueof$$ $${\int }_0^{+\mathrm{\infty }}\frac{e^{-at}-e^{-bt}}{t}cos\left(xt\right)dt.$$

Commented byabdo imad last updated on 19/Feb/18

$$letputF\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-at}-e^{-bt}}{t}cos\left(xt\right)dtafterverifyingtbat$$ $$Fisderivablewehave$$ $$F^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial x}\left(\frac{e^{-at}-e^{-bt}}{t}cos\left(xt\right)\right)dt$$ $$=-{\int }_0^{\mathrm{\infty }}(e^{-at}-e^{-bt})sin\left(xt\right)dt=Im\left({\int }_0^{\mathrm{\infty }}(e^{-bt}-e^{-at})e^{ixt}dt\right)$$ $$=Im({\int }_0^{\mathrm{\infty }}{e}^{(-b+ix)t}-e^{(-a+ix)t})dtbut$$ $${\int }_0^{\mathrm{\infty }}{e}^{(-b+ix)t}dt=\frac{1}{-b+ix}{\left[e^{(-\mathit{b}+\mathit{i}\mathit{x})\mathit{t}}\right]}_{\mathit{t}=0}^{\mathrm{\infty }}=\frac{-1}{-b+ix}=\frac{1}{b-ix}$$ $$alsowehave{\int }_0^{\mathrm{\infty }}{e}^{(-a+ix)t}dt=\frac{1}{a-ix}\Rightarrow$$ $$\frac{dF}{dx}\left(x\right)=\frac{1}{b-ix}-\frac{1}{a-ix}=\frac{b+ix}{b^2+x^2}-\frac{a+ix}{a^2+x^2}$$ $$=\frac{b}{b^2+x^2}-\frac{a}{a^2+x^2}+ix(\frac{1}{b^2+x^2}-\frac{1}{a^2+x^2})\Rightarrow$$ $$\frac{dF}{dx}\left(x\right)=x\frac{a^2+x^2-b^2-x^2}{(a^2+x^2)(b^2+x^2)}=\frac{(a^2-b^2)x}{(a^2+x^2)(b^2+x^2)}\Rightarrow$$ $$F\left(x\right)=\int \frac{x}{b^2+x^2}dx-\int \frac{x}{a^2+x^2}dx+\lambda$$ $$=\frac{1}{2}ln(b^2+x^2)-\frac{1}{2}ln(a^2+x^2)+\lambda$$ $$F\left(x\right)=\frac{1}{2}ln\left(\frac{x^2+b^2}{x^2+a^2}\right)+\lambda$$ $$\lambda ={lim}_{x\to 0}(F\left(x\right)-\frac{1}{2}ln\left(\frac{x^2+b^2}{x^2+a^2}\right))=F\left(o\right)-ln\left(\frac{b}{a}\right)$$ $$\lambda ={\int }_0^{\mathrm{\infty }}\frac{e^{-at}-e^{-bt}}{t}dt-ln\left(\frac{b}{a}\right)letput$$ $$I\left(\xi \right)={\int }_{\xi }^{+\mathrm{\infty }}\frac{e^{-at}-e^{-bt}}{t}dt={\int }_{\xi }^{\mathrm{\infty }}\frac{e^{-at}}{t}dt-{\int }_{\xi }^{\mathrm{\infty }}\frac{e^{-bt}}{t}dtch.at=u$$ $$give{\int }_{\xi }^{\mathrm{\infty }}\frac{e^{-at}}{t}dt={\int }_{a\xi }^{+\mathrm{\infty }}\frac{e^{-u}}{\frac{u}{a}}\frac{du}{a}={\int }_{a\xi }^{+\mathrm{\infty }}\frac{e^{-u}}{u}du\Rightarrow$$ $$I\left(\xi \right)={\int }_{a\xi }^{+\mathrm{\infty }}\frac{e^{-u}}{u}du-{\int }_{b\xi }^{+\mathrm{\infty }}\frac{e^{-u}}{u}du={\int }_{a\xi }^{b\xi }\frac{e^{-u}}{u}dubut$$ $$\mathrm{\exists }{c}_{}\in ]a\xi ,b\xi [/I\left(\xi \right)=e^{-\xi }{\int }_{a\xi }^{b\xi }\frac{du}{u}=e^{-\xi }ln\left(\frac{b}{a}\right)\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}\frac{e^{-at}-e^{-bt}}{t}dt={lim}_{\xi \to 0}I\left(\xi \right)=ln\left(\frac{b}{a}\right)\Rightarrow \lambda =0and$$ $$F\left(x\right)=\frac{1}{2}ln\left(\frac{x^2+b^2}{x^2+a^2}\right).$$

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