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Question Number 72359 by aliesam last updated on 27/Oct/19

Commented by mind is power last updated on 27/Oct/19

$$\mathrm{nice}\mathrm{one}\mathrm{sir}\mathrm{thanx}$$

Commented by aliesam last updated on 27/Oct/19

$$thankyousir$$$$$$

Answered by mind is power last updated on 27/Oct/19

$$$$$$\mathrm{E}={\int }_0^{+\mathrm{\infty }}\frac{(1+\mathrm{x}+{\mathrm{x}}^2)\log \left(\mathrm{x}\right)}{1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4}\mathrm{dx}+{\int }_0^{+\mathrm{\infty }}\frac{\mathrm{xlog}\left(2\right)+{\mathrm{x}}^2\log \left(3\right)}{1+{\mathrm{x}}^2+{\mathrm{x}}^3+\mathrm{x}+{\mathrm{x}}^4}\mathrm{dx}$$$$1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4=({\mathrm{x}}^2-2\cos \left(\frac{2\pi }{5}\right)\mathrm{x}+1)({\mathrm{x}}^2-2\cos \left(\frac{4\pi }{5}\right)\mathrm{x}+1)$$$$\cos \left(\frac{2\pi }{5}\right)=\frac{-1+\sqrt{5}}{4}$$$$\cos \left(\frac{4\pi }{5}\right)=\frac{2.(6-2\sqrt{5})}{16}-1=\frac{-4-4\sqrt{5}}{16}=-\frac{\sqrt{5}+1}{4}$$$$1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4=({\mathrm{x}}^2-\left(\frac{-1+\sqrt{5}}{2}\right)\mathrm{x}+1)({\mathrm{x}}^2+\left(\frac{1+\sqrt{5}}{2}\right)\mathrm{x}+1)$$$${\int }_0^{+\mathrm{\infty }}\frac{(1+\mathrm{x}+{\mathrm{x}}^2)\log \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$$$\mathrm{let}\mathrm{t}=\frac{1}{\mathrm{x}}$$$${\int }_0^{+\mathrm{\infty }}\frac{(1+\mathrm{x}+{\mathrm{x}}^2)\log \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}={\int }_{\mathrm{\infty }}^0\frac{(1+\frac{1}{\mathrm{t}}+\frac{1}{{\mathrm{t}}^2})(-\log \left(\mathrm{t}\right))}{(\frac{1}{{\mathrm{t}}^4}+\frac{1}{{\mathrm{t}}^3}+\frac{1}{{\mathrm{t}}^2}+\frac{1}{{\mathrm{t}}^{}}+1)}.\frac{-\mathrm{dt}}{{\mathrm{t}}^2}$$$$={\int }_{\mathrm{\infty }}^0\frac{({\mathrm{t}}^2+\mathrm{t}+1)\log \left(\mathrm{t}\right)}{(1+\mathrm{t}+{\mathrm{t}}^2+{\mathrm{t}}^3+{\mathrm{t}}^4)}=-{\int }_0^{+\mathrm{\infty }}\frac{(1+\mathrm{x}+{\mathrm{x}}^2)\log \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}\Rightarrow$$$$2{\int }_0^{+\mathrm{\infty }}\frac{(1+\mathrm{x}+{\mathrm{x}}^2)\log \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}=0\Rightarrow {\int }_0^{+\mathrm{\infty }}\frac{(1+\mathrm{x}+{\mathrm{x}}^2)\log \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}=0$$$$\iff$$$$\mathrm{E}={\int }_0^{+\mathrm{\infty }}\frac{\mathrm{xlog}\left(2\right)+{\mathrm{x}}^2\log \left(3\right)}{1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4}={\int }_0^{+\mathrm{\infty }}\frac{\mathrm{xlog}\left(2\right)+{\mathrm{x}}^2\log \left(3\right)}{({\mathrm{x}}^2-\left(\frac{-1+\sqrt{5}}{2}\right)\mathrm{x}+1)({\mathrm{x}}^2+\left(\frac{1+\sqrt{5}}{2}\right)\mathrm{x}+1)}$$$$\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2-\left(\frac{-1+\sqrt{5}}{2}\right)\mathrm{x}+1}+\frac{\mathrm{cx}+\mathrm{d}}{({\mathrm{x}}^2+\left(\frac{1+\sqrt{5}}{2}\right)\mathrm{x}+1}$$$$\mathrm{a}+\mathrm{c}=0$$$$\mathrm{b}+\mathrm{d}=0$$$$(\mathrm{a}-\mathrm{c})\frac{\sqrt{5}}{2}=\log \left(3\right)$$$$(\mathrm{b}-\mathrm{d})\frac{\sqrt{5}}{2}=\log \left(2\right)$$$$\mathrm{a}=\frac{\log \left(3\right)}{\sqrt{5}},\mathrm{c}=-\frac{\log \left(3\right)}{\sqrt{5}}$$$$\mathrm{b}=\frac{\log \left(2\right)}{\sqrt{5}},\mathrm{d}=\frac{-\log \left(2\right)}{\sqrt{5}}$$$$\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2-\left(\frac{-1+\sqrt{5}}{2}\right)\mathrm{x}+1}+\frac{\mathrm{cx}+\mathrm{d}}{({\mathrm{x}}^2+\left(\frac{1+\sqrt{5}}{2}\right)\mathrm{x}+1}=\frac{1}{\sqrt{5}}(\frac{\mathrm{xlog}\left(3\right)+\log \left(2\right)}{{\mathrm{x}}^2-\left(\frac{-1+\sqrt{5}}{2}\right)\mathrm{x}+1}+\frac{-\mathrm{xlog}\left(3\right)-\log \left(2\right)}{{\mathrm{x}}^2+\left(\frac{1+\sqrt{5}}{2}\right)\mathrm{x}+1})$$$$\mathrm{now}\mathrm{just}\mathrm{integrat}\mathrm{bee}\mathrm{continued}$$$$$$$$$$

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