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Question Number 163954 by mnjuly1970 last updated on 12/Jan/22

Answered by mathmax by abdo last updated on 13/Jan/22

$$\mathrm{\Psi }=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}}{\mathrm{sh}\left(2\mathrm{x}\right)}\mathrm{dx}{=}_{2\mathrm{x}=\mathrm{t}}2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2{\mathrm{e}}^{\frac{\mathrm{t}}{2}}}{4\mathrm{sh}\left(\mathrm{t}\right)}\frac{\mathrm{dt}}{2}=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2{\mathrm{e}}^{\frac{\mathrm{t}}{2}}}{\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}}\mathrm{dt}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2{\mathrm{e}}^{\frac{\mathrm{t}}{2}}}{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2{\mathrm{e}}^{-\mathrm{t}+\frac{\mathrm{t}}{2}}}{1-{\mathrm{e}}^{-2\mathrm{t}}}\mathrm{dt}$$$$\frac{1}{2}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{-\frac{\mathrm{t}}{2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{nt}}\mathrm{dt}$$$$=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{-\frac{\mathrm{t}}{2}-2\mathrm{nt}}\mathrm{dt}$$$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{-(2\mathrm{n}+\frac{1}{2})\mathrm{t}}\mathrm{dt}{=}_{(2\mathrm{n}+\frac{1}{2})\mathrm{t}=\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^2}{{(2\mathrm{n}+\frac{1}{2})}^2}{\mathrm{e}}^{-\mathrm{z}}\frac{\mathrm{dz}}{(2\mathrm{n}+\frac{1}{2})}$$$$=\frac{1}{{(2\mathrm{n}+\frac{1}{2})}^3}{\int }_0^{\mathrm{\infty }}{\mathrm{z}}^2{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}=\frac{\mathrm{\Gamma }\left(3\right)}{{(2\mathrm{n}+\frac{1}{2})}^3}\Rightarrow$$$$\mathrm{\Psi }=\frac{2!}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(2\mathrm{n}+\frac{1}{2})}^3}=8\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(4\mathrm{n}+1)}^3}=\frac{8}{4^3}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+\frac{1}{4})}^3}$$$$\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{this}\mathrm{sum}\mathrm{using}\mathrm{digamma}\mathrm{function}....$$

Answered by Lordose last updated on 13/Jan/22

$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}}{\sinh \left(\mathrm{x}\right)\cosh \left(\mathrm{x}\right)\mathrm{dx}}=2{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\mathrm{csch}\left(2\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{I}\stackrel{\mathrm{x}=\frac{\mathrm{x}}{2}}{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2{\mathrm{e}}^{\frac{\mathrm{x}}{2}}}{{\mathrm{e}}^{\mathrm{x}}(1-{\mathrm{e}}^{-2\mathrm{x}})}\mathrm{dx}\stackrel{\mathrm{x}=-\ln \left(\mathrm{x}\right)}{=}\frac{1}{2}{\int }_0^1\frac{{\mathrm{x}}^{-\frac{1}{2}}{\ln }^2\left(\mathrm{x}\right)}{1-{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{I}=\frac{1}{2}{\int }_0^1\frac{{\mathrm{x}}^{-\frac{1}{2}}{\ln }^2\left(\mathrm{x}\right)}{1-{\mathrm{x}}^2}\mathrm{dx}\stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{2}}}{=}\frac{1}{16}{\int }_0^1\frac{{\mathrm{x}}^{\frac{1}{4}-1}{\ln }^2\left(\mathrm{x}\right)}{1-\mathrm{x}}\mathrm{dx}$$$$-{\int }_0^1\frac{{\mathbf{t}}^{\mathbf{x}-1}{\mathbf{ln}}^2\left(\mathbf{x}\right)}{1-\mathbf{t}}\mathbf{dx}={\mathit{\psi }}^{\left(2\right)}\left(\mathbf{x}\right)$$$$\mathrm{I}=-\frac{1}{16}{\mathit{\psi }}^{\left(2\right)}\left(\frac{1}{4}\right)=\frac{7}{2}\mathit{\zeta }\left(3\right)+\frac{{\mathit{\pi }}^3}{8}$$$$\mathit{\alpha }=\frac{7}{2},\mathit{\beta }=\frac{1}{8}$$$$\mathit{\varnothing }\mathbf{sE}$$

Commented by mnjuly1970 last updated on 13/Jan/22

$$mercrysirlordos$$

Answered by mindispower last updated on 13/Jan/22

$$={\int }_0^{\mathrm{\infty }}\frac{4x^2{e}^{-x}}{(1+e^{-2x})(1-e^{-2x})}dx$$$$=2({\int }_0^{\mathrm{\infty }}\frac{x^2{e}^{-x}}{1+e^{-2x}}dx+{\int }_0^{\mathrm{\infty }}\frac{x^2{e}^{-x}}{1-e^{-2x}})$$$$=2\sum _{n⩾0}({\int }_0^{\mathrm{\infty }}{x}^2{(-1)}^n{e}^{-(1+2n)x}dx+{\int }_0^{\mathrm{\infty }}{x}^2{e}^{-(1+2n)dx}$$$$=2\sum _{n⩾0}\frac{{(-1)}^n}{{(2n+1)}^3}\mathrm{\Gamma }\left(3\right)+2\sum _{n⩾0}\frac{1}{{(1+2n)}^3}\mathrm{\Gamma }\left(3\right)$$$$=4\beta \left(3\right)+4(\zeta \left(3\right)-\frac{\zeta \left(3\right)}{8})$$$$=4\beta \left(3\right)+\frac{7}{2}\zeta \left(3\right)=\frac{{\pi }^3}{8}+\frac{7}{2}\zeta \left(3\right)$$$$a+b=\frac{29}{8}$$

Commented by mnjuly1970 last updated on 13/Jan/22

$$verynicesolutionasalways$$$$sirpower...$$

Commented by mindispower last updated on 14/Jan/22

$$withePleasurniceday$$

Commented by mnjuly1970 last updated on 14/Jan/22

Answered by mnjuly1970 last updated on 13/Jan/22

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