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Question Number 29971 by abdo imad last updated on 14/Feb/18

$$findJ\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{x+e^t}?.$$

Commented by abdo imad last updated on 16/Feb/18

$$J\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{e^t(1+x{e}^{-t})}={\int }_0^{\mathrm{\infty }}{e}^{-t}\left(\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{-nt}\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{x}^n{\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t}dtthe(n+1)t=ugive$$$${\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t}dt={\int }_0^{\mathrm{\infty }}{e}^{-u}\frac{du}{n+1}=\frac{1}{n+1}{[-e^{-u}]}_0^{+\mathrm{\infty }}=\frac{1}{n+1}$$$$J\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n+1}\Rightarrow xJ\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{n+1}letderivate$$$$J\left(x\right)+{xJ}^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}soJissolutionofthe$$$$d.ey+{xy}^{{}^{\prime }}=\frac{1}{1-x}eh\Rightarrow {xy}^{{}^{\prime }}=-y\Rightarrow \frac{y^{{}^{\prime }}}{y}=\frac{-1}{x}\Rightarrow$$$$ln\mid y\mid =-ln\mid x\mid +c\Rightarrow y=\frac{\lambda }{x}mvcmethodgive$$$$y^{{}^{\prime }}=\frac{{\lambda }^{{}^{\prime }}x-\lambda }{x^2}\Rightarrow \frac{\lambda }{x}+{\lambda }^{{}^{\prime }}-\frac{\lambda }{x}=\frac{1}{1-x}\Rightarrow \lambda \left(x\right)=\int \frac{dx}{1-x}+\kappa$$$$=-ln\mid 1-x\mid +kbutk=\lambda \left(0\right)=0$$$$y\left(x\right)=-\frac{1}{x}ln\mid 1-x\mid \Rightarrow J\left(x\right)=-\frac{1}{x}ln\mid 1-x\mid .$$

Commented by abdo imad last updated on 16/Feb/18

$$theQ.isfindJ\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{x+e^t}with\mid x\mid <1.$$

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