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Question Number 108417 by mathdave last updated on 16/Aug/20

Answered by Aziztisffola last updated on 16/Aug/20

$$\mathrm{t}=2\mathrm{x}\Rightarrow \mathrm{dt}=2\mathrm{dx}$$$${\int }_0^{\mathrm{\infty }}\frac{2\mathrm{x}}{{\mathrm{e}}^{2\mathrm{x}}-1}\mathrm{dx}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{{\mathrm{e}}^{\mathrm{t}}-1}\mathrm{dt}$$$$=\frac{1}{2}\mathrm{\Gamma }\left(2\right)\zeta \left(2\right)=\frac{1}{2}\times 1\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}$$

Answered by mnjuly1970 last updated on 16/Aug/20

$$\mathrm{ans}:2x=t$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{2x}{e^{2x}-1}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t}{e^t-1}dt$$$$\mathrm{\Omega }=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{s-1}}{{\mathrm{e}}^t-1}dt{\mid }_{s=2}=\frac{1}{2}\zeta \left(2\right)\mathrm{\Gamma }\left(2\right)=\frac{{\pi }^2}{12}\mathrm{♣}$$$$...\mathrm{♣}\mathcal{M}.\mathcal{N}azerian\mathrm{♣}...$$$$$$

Answered by mathmax by abdo last updated on 16/Aug/20

$$\mathrm{I}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{e}}^{2\mathrm{x}}-1}\mathrm{dx}=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{xe}}^{-2\mathrm{x}}}{1-{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}=2{\int }_0^{\mathrm{\infty }}{\mathrm{xe}}^{-2\mathrm{x}}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{nx}}\right)\mathrm{dx}$$$$=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\mathrm{x}{\mathrm{e}}^{-2(\mathrm{n}+1)\mathrm{x}}\mathrm{dx}{=}_{2(\mathrm{n}+1)\mathrm{x}=\mathrm{t}}2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{2(\mathrm{n}+1)}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{2(\mathrm{n}+1)}$$$$=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2}{\int }_0^{\mathrm{\infty }}{\mathrm{te}}^{-\mathrm{t}}\mathrm{dt}=\frac{1}{2}\xi \left(2\right).\mathrm{\Gamma }\left(2\right)$$$$=\frac{{\pi }^2}{12}$$

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