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Question Number 27345 by abdo imad last updated on 05/Jan/18

$$provethat{\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{e^t-1}dt=\xi \left(x\right)\mathrm{\Gamma }\left(x\right)$$$$with\xi \left(x\right)=\sum _{n=1}^{\propto }\frac{1}{n^x}and\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$$$(x>1)$$

Commented by abdo imad last updated on 07/Jan/18

$${\int }_0^{\propto }\frac{t^{x-1}}{e^t-1}dt={\int }_0^{\propto }\frac{e^{-t}{t}^{x-1}}{1-e^{-t}}dt$$$$={\int }_0^{\propto }\left(\sum _{n=0}^{\propto }{e}^{-nt}\right)e^{-t}{t}^{x-1}dt$$$$=\sum _{n=0}^{\propto }\left({\int }_0^{\propto }{e}^{-(n+1)t}{t}^{x-1}dt\right)thech.(n+1)t=ugive$$$${\int }_0^{\propto }{e}^{-(n+1)t}{t}^{x-1}dt=\frac{1}{n+1}{\int }_0^{\propto }{e}^{-u}{\left(\frac{u}{n+1}\right)}^{x-1}du$$$$=\frac{1}{{(n+1)}^x}{\int }_0^{\propto }{e}^{-u}{u}^{x-1}du=\frac{\mathrm{\Gamma }\left(x\right)}{{(n+1)}^x}$$$$\Rightarrow {\int }_0^{\propto }\frac{t^{x-1}}{e^t-1}dt=\sum _{n=0}^{\propto }\left(\frac{\mathrm{\Gamma }\left(x\right)}{{(n+1)}^x}\right)=\left(\sum _{n=1}^{\propto }\frac{1}{n^x}\right)\mathrm{\Gamma }\left(x\right)$$$$=\xi \left(x\right).\mathrm{\Gamma }\left(x\right).$$$$$$

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