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Question Number 45045 by maxmathsup by imad last updated on 07/Oct/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{t^3}{1+e^t}dt.$$

Answered by maxmathsup by imad last updated on 09/Oct/18

$$letA={\int }_0^{\mathrm{\infty }}\frac{t^3}{1+e^t}dtwehaveA={\int }_0^{\mathrm{\infty }}\frac{t^3{e}^{-t}}{1+e^{-t}}dt$$$$={\int }_0^{\mathrm{\infty }}{t}^3{e}^{-t}\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-nt}\right)dt=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{t}^3{e}^{-(n+1)t}dt$$$$changement(n+1)t=xgive$$$$A=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{\left(\frac{x}{n+1}\right)}^3{e}^{-x}\frac{dx}{(n+1)}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^4}{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-x}dtletrememberthat\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt(x>0)\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{x}^3{e}^{-x}dx={\int }_0^{\mathrm{\infty }}{x}^{4-1}{e}^{-x}dx=\mathrm{\Gamma }\left(4\right)=(4-1)!=3!=6\Rightarrow$$$$A=6\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^4}=6\delta \left(4\right)with\delta \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^x}(x>0)letfind\delta \left(x\right)$$$$intermsof\xi \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}wehave\delta \left(x\right)=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^x}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}$$$$=-2^{-x}\xi \left(x\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}but\xi \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^x}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}$$$$=2^{-x}\xi \left(x\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}=(1-2^{-x})\xi \left(x\right)\Rightarrow$$$$\delta \left(x\right)=-2^{-x}\xi \left(x\right)+(1-2^{-x})\xi \left(x\right)=(1-2^{1-x})\xi \left(x\right)\Rightarrow$$$$\delta \left(4\right)=(1-2^{-3})\xi \left(4\right)=(1-\frac{1}{8})\frac{{\pi }^4}{90}=\frac{7}{8}\frac{{\pi }^4}{90}\Rightarrow A=6.\frac{7}{8}.\frac{{\pi }^4}{90}\Rightarrow$$$$=\frac{2.3.7}{2.4}.\frac{{\pi }^4}{3.30}=\frac{7{\pi }^4}{120}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{t^3}{1+e^t}dt=\frac{7{\pi }^4}{120}★$$

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