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Question Number 119989 by mnjuly1970 last updated on 28/Oct/20

Answered by mathmax by abdo last updated on 28/Oct/20

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{xdx}}{{2\mathrm{e}}^{\mathrm{x}}-1}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{xe}}^{-\mathrm{x}}\mathrm{dx}}{2-{\mathrm{e}}^{-\mathrm{x}}}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{xe}}^{-\mathrm{x}}}{1-2^{-1}{\mathrm{e}}^{-\mathrm{x}}}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{\mathrm{xe}}^{-\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}^{\mathrm{n}}=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}{\int }_0^{\mathrm{\infty }}{\mathrm{xe}}^{-(\mathrm{n}+1)\mathrm{x}}\mathrm{dx}$$$${=}_{(\mathrm{n}+1)\mathrm{x}=\mathrm{t}}\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{n}+1}$$$$=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2{2}^{\mathrm{n}}}{\int }_0^{\mathrm{\infty }}\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2{2}^{\mathrm{n}}}\mathrm{\Gamma }\left(2\right)(\mathrm{\Gamma }\left(2\right)=1)$$$$=\frac{1}{2}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2{2}^{\mathrm{n}-1}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2\times 2^{\mathrm{n}}}.....\mathrm{be}\mathrm{continued}...$$

Commented by mnjuly1970 last updated on 29/Oct/20

$$grateful$$$$mrmax.excellent...$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n{n}^2}={li}_2\left(\frac{1}{2}\right)✓$$

Commented by Bird last updated on 29/Oct/20

$$youarewelcome$$

Answered by mindispower last updated on 30/Oct/20

$${\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(x\right)ln(x+1)}{x(x+1)}dx$$$$={\int }_0^1\frac{{ln}^2\left(x\right)ln(1+x)}{x(x+1)}+{\int }_1^{\mathrm{\infty }}\frac{{ln}^2\left(x\right)ln(1+x)}{x(x+1)}{dx}_{x\to \frac{1}{x}}$$$$={\int }_0^1\frac{{ln}^2\left(x\right)ln(1+x)(x+1-x)}{x(x+1)}dx+{\int }_0^1\frac{{ln}^2\left(x\right)ln(1+\frac{1}{x})}{1+x}dx$$$$={\int }_0^1\frac{{ln}^2\left(x\right)ln(1+x)}{x}dx-{\int }_0^1\frac{{ln}^2\left(x\right)ln(1+x)}{1+x}dx$$$$+{\int }_0^1\frac{{ln}^2\left(x\right)ln(1+x)}{1+x}dx-{\int }_0^1\frac{{ln}^3\left(x\right)}{1+x}dx$$$$={\int }_0^1\frac{{ln}^2\left(x\right)ln(1+x)dx}{x}-{\int }_0^1\frac{{ln}^3\left(x\right)}{1+x}$$$$={\left[\frac{{ln}^3\left(x\right)ln(1+x)}{3}\right]}_0^1-{\int }_0^1\frac{{ln}^3\left(x\right)}{3(1+x)}dx-{\int }_0^1\frac{{ln}^3\left(x\right)}{1+x}dx$$$$=\frac{4}{3}{\int }_0^1\frac{{(-ln\left(x\right))}^3}{1+x}{dx}_{x=e^{-t}}$$$$=\frac{4}{3}{\int }_0^{\mathrm{\infty }}\frac{t^3}{1+e^{-t}}{e}^{-t}dt=\frac{4}{3}\sum _{k⩾0}{\int }_0^{\mathrm{\infty }}{t}^3{e}^{-(1+k)t}dt$$$$=\frac{4}{3}\sum _{k⩾0}{\int }_0^{\mathrm{\infty }}{(-1)}^k\frac{t^3{e}^{-t}dt}{{(1+k)}^4}$$$$=\frac{4}{3}\sum _{k⩾0}\frac{{(-1)}^k}{{(1+k)}^4}\mathrm{\Gamma }\left(4\right)=8\sum _{k⩾0}\frac{{(-1)}^k}{{(1+k)}^4}$$$$8(\sum _{k⩾0}\frac{1}{{(2k+1)}^4}-\sum _{k⩾0}\frac{1}{{(2k+2)}^4})=8(\zeta \left(4\right)-\frac{1}{16}\zeta \left(4\right)-\frac{\zeta \left(4\right)}{16})$$$$=8.\frac{14}{16}\zeta \left(4\right)=7\zeta \left(5\right)$$$$$$

Commented by mnjuly1970 last updated on 30/Oct/20

$$bravomrmindspower$$$$veryniceasalways..$$$$thankyou...$$

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