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Question Number 130759 by mnjuly1970 last updated on 28/Jan/21

$$...nicecalculus...$$$$find:\varphi ={\int }_0^{\mathrm{\infty }}(x^{2}ln(1+e^x)-x^3)dx$$$$$$

Answered by mnjuly1970 last updated on 28/Jan/21

$$ans:$$$$\varphi ={\int }_0^{\mathrm{\infty }}{x}^{2}ln\left(\frac{1+e^x}{e^x}\right)dx$$$$={\int }_0^{\mathrm{\infty }}{x}^{2}ln(1+e^{-x})dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{(n-1)}\frac{1}{n}{\int }_0^{\mathrm{\infty }}{x}^2{e}^{-nx}dx$$$$\stackrel{nx=y}{=}\underset{n=1}{\sum ^{\mathrm{\infty }}}\left[{(-1)}^{(n-1)}\frac{1}{n^4}{\int }_0^{\mathrm{\infty }}{y}^2{e}^{-y}dy\right]$$$$=\mathrm{\Gamma }\left(3\right)\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n^4}=2[\frac{1}{1^4}-\frac{1}{2^4}+\frac{1}{3^4}-\frac{1}{4^4}+..]$$$$=2(\zeta \left(4\right)-\frac{2}{2^4}\left(\zeta \left(4\right)\right))=2(\frac{{\pi }^4}{90}-\frac{{\pi }^4}{8\ast 90})$$$$=\frac{{\pi }^4}{45}(1-\frac{1}{8})=\frac{7}{8}\ast \frac{{\pi }^4}{45}=\frac{7{\pi }^4}{360}=...$$$$correctorno???$$

Answered by Dwaipayan Shikari last updated on 28/Jan/21

$${\int }_0^{\mathrm{\infty }}{x}^{2}log(1+e^{-x})dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n}{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{2}dx=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n^4}=\frac{{\pi }^4}{45}(1-\frac{1}{2^3})=\frac{7{\pi }^4}{360}$$

Commented by mnjuly1970 last updated on 28/Jan/21

$$niceverynice::$$$$\eta \left(s\right)=(1-2^{1-s})\zeta \left(s\right)$$$$\eta \left(4\right)=(1-\frac{1}{8})\zeta \left(4\right)=\frac{7}{8}\ast \frac{{\pi }^4}{90}$$$$=\frac{7{\pi }^4}{720}\Rightarrow ans:=2\left(\frac{7{\pi }^4}{720}\right)=\frac{7{\pi }^4}{360}$$$$$$

Answered by Lordose last updated on 28/Jan/21

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}({\mathrm{x}}^2\ln (1+{\mathrm{e}}^{\mathrm{x}})-{\mathrm{x}}^3)\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^2\left(\ln (1+{\mathrm{e}}^{-\mathrm{x}})\right)\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathrm{u}={\mathrm{e}}^{-\mathrm{x}}}{=}{\int }_0^1\frac{1}{\mathrm{u}}{(-\mathrm{lnu})}^2\ln (1+\mathrm{u})\mathrm{du}={\int }_0^1{\mathrm{u}}^{-1}\ln (1+\mathrm{u}){\ln }^2\left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{\Omega }=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{u}}^{\mathrm{n}-1}{\ln }^2\left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{\Omega }\left(\mathrm{n}\right)={\int }_0^1{\mathrm{u}}^{\mathrm{n}}\mathrm{du}=\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}=\frac{1}{\mathrm{n}+1}$$$${\mathrm{\Omega }}^{\prime }\left(\mathrm{n}\right)={\int }_0^1{\mathrm{u}}^{\mathrm{n}}\ln \left(\mathrm{u}\right)\mathrm{du}=-\frac{1}{{(\mathrm{n}+1)}^2}$$$${\mathrm{\Omega }}^{″}\left(\mathrm{n}\right)={\int }_0^1{\mathrm{u}}^{\mathrm{n}}{\ln }^2\left(\mathrm{u}\right)\mathrm{du}=\frac{1}{{(\mathrm{n}+1)}^3}$$$${\mathrm{\Omega }}^{{}^{″}}(\mathrm{n}-1)=\frac{1}{{\mathrm{n}}^3}$$$$\mathrm{\Omega }=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^4}=\frac{7{\pi }^4}{720}$$

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