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Question Number 53783 by maxmathsup by imad last updated on 25/Jan/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{t^2}{e^t-1}dtintermsof\xi \left(3\right)$$

Answered by Smail last updated on 26/Jan/19

$$A{\int }_0^{\mathrm{\infty }}\frac{t^2}{e^t-1}dt={\int }_0^{\mathrm{\infty }}\frac{t^2{e}^{-t}}{1-e^{-t}}dt$$$$={\int }_0^{\mathrm{\infty }}{t}^2{e}^{-t}\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-nt}dt$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{t}^2{e}^{-(n+1)t}dt$$$$byparts$$$$u=t^2\Rightarrow u^{\prime }=2t$$$$v^{\prime }=e^{-(n+1)t}\Rightarrow v=\frac{-1}{n+1}{e}^{-(n+1)t}$$$$A=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{n+1}{\int }_0^{\mathrm{\infty }}{te}^{-(n+1)t}dtwith({\left[t^2{e}^{-(n+1)t}\right]}_0^{\mathrm{\infty }}=0)$$$$byparts$$$$A=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{{(n+1)}^2}{\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t}dt$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{{(n+1)}^3}{\left[e^{-(n+1)t}\right]}_0^{\mathrm{\infty }}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1)}^3}$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}=2\xi \left(3\right)$$

Commented by maxmathsup by imad last updated on 26/Jan/19

$$thankyousir$$

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