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Question Number 33353 by caravan msup abdo. last updated on 15/Apr/18

$$letx\in ]1,+\mathrm{\infty }[and\lambda \in [-1,1]$$$$givetheintegral{\int }_0^{\mathrm{\infty }}\frac{t^{x-1}{e}^{-t}}{1-\lambda e^{-t}}dt$$$$atformofserie.$$

Commented by math khazana by abdo last updated on 18/Apr/18

$$letputA\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{t^{x-1}{e}^{-t}}{1-\lambda e^{-t}}dt$$$$A\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}\left(\sum _{n=0}^{\mathrm{\infty }}{\lambda }^n{e}^{-nt}\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{\lambda }^n{\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-(n+1)t}dtch.(n+1)t=ugive$$$$A\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{\lambda }^n{\int }_0^{\mathrm{\infty }}{\left(\frac{u}{n+1}\right)}^{x-1}{e}^{-u}\frac{du}{n+1}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{\lambda }^n}{{(n+1)}^x}{\int }_0^{\mathrm{\infty }}{u}^{x-1}{e}^{-u}du$$$$=\mathrm{\Gamma }\left(x\right)\sum _{n=0}^{\mathrm{\infty }}\frac{{\lambda }^n}{{(n+1)}^x}.$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{u}^{x-1}{e}^{-u}du$$

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