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Question Number 114135 by mnjuly1970 last updated on 17/Sep/20

$$....\mathcal{A}dvancedmathematics...$$$$i::provethat:\mathrm{\Omega }=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{1}{{(x^2-x+1)}^2\sqrt{x}}dx=1$$$$ii::evaluate::\mathrm{\Phi }={\int }_0^1{x}^{2}ln\left(x\right)ln(1-x)dx=???$$$$....m.n.july.1970....$$$$$$

Answered by MJS_new last updated on 17/Sep/20

$$\int \frac{dx}{{(x^2-x+1)}^2\sqrt{x}}=$$$$[t=\sqrt{x}\to dx=2\sqrt{x}dt]$$$$=2\int \frac{dt}{{(t^4-t^2+1)}^2}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{t(t^2+1)}{3(t^4-t^2+1)}+\frac{1}{3}\int \frac{t^2+5}{t^4-t^2+1}dt$$$$\frac{1}{3}\int \frac{t^2+5}{t^4-t^2+1}dt=$$$$=\frac{1}{18}\int (\frac{4\sqrt{3}t+15}{t^2+\sqrt{3}t+1}-\frac{4\sqrt{3}t-15}{t^2-\sqrt{3}t+1})dt=$$$$\left[\mathrm{using}\mathrm{formula}\right]$$$$=\frac{\sqrt{3}}{9}\ln \frac{t^2+\sqrt{3}t+1}{t^2-\sqrt{3}t+1}+\arctan (2t-\sqrt{3})+\arctan (2t+\sqrt{3})$$$$\Rightarrow$$$$\int \frac{dx}{{(x^2-x+1)}^2\sqrt{x}}=$$$$=\frac{(x+1)\sqrt{x}}{3(x^2-x+1)}+\ln \frac{x+1+\sqrt{3x}}{x+1-\sqrt{3x}}+\arctan (2\sqrt{x}-\sqrt{3})+\arctan (2\sqrt{x}+\sqrt{3})+C$$$$\Rightarrow \frac{1}{\pi }\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x^2-x+1)}^2\sqrt{x}}=1$$

Commented by mnjuly1970 last updated on 17/Sep/20

$$thankyousomuchmr$$

Answered by mathmax by abdo last updated on 18/Sep/20

$$\mathrm{let}\mathrm{prove}\mathrm{thst}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2\sqrt{\mathrm{x}}}=\pi$$$$\mathrm{changement}\sqrt{\mathrm{x}}=\mathrm{t}\mathrm{give}\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{tdt}}{{({\mathrm{t}}^4-{\mathrm{t}}^2+1)}^2\mathrm{t}}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{({\mathrm{t}}^4-{\mathrm{t}}^2+1)}^2}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dt}}{{({\mathrm{t}}^4-{\mathrm{t}}^2+1)}^2}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{1}{{({\mathrm{z}}^4-{\mathrm{z}}^2+1)}^2}$$$$\mathrm{poles}\mathrm{of}\phi ?{\mathrm{z}}^4-{\mathrm{z}}^2+1=0\Rightarrow {\mathrm{u}}^2-\mathrm{u}+1=0(\mathrm{u}={\mathrm{z}}^2)$$$$\mathrm{\Delta }=-3\Rightarrow {\mathrm{u}}_1=\frac{1+3\mathrm{i}}{2}={\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\mathrm{and}{\mathrm{u}}_2=\frac{1-3\mathrm{i}}{2}={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}$$$$\Rightarrow {\mathrm{z}}^4-{\mathrm{z}}^2+1=({\mathrm{z}}^2-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})=(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{6}})(\mathrm{z}+{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{6}})$$$$\Rightarrow \phi \left(\mathrm{z}\right)=\frac{1}{{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^2{(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^2{(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{6}})}^2{(\mathrm{z}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{6}})}^2}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})+\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{6}})\}$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})={\lim }_{\mathrm{z}\to {\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^2\phi \left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to {\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}}{\left\{\frac{1}{{(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^2{({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})}^2}\right\}}^{\left(1\right)}$$$$=-{\lim }_{\mathrm{z}\to {\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}}\frac{2(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}){({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})}^2+4\mathrm{z}({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}){(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^2}{{(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^4{({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})}^4}$$$$=-{2\lim }_{\mathrm{z}\to {\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}}\frac{({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})+2\mathrm{z}(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}{{(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})}^3{({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})}^3}....\mathrm{be}\mathrm{continued}....$$

Commented by mathmax by abdo last updated on 18/Sep/20

$$\mathrm{let}\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4-{\mathrm{x}}^2+1)}^2}\mathrm{let}\mathrm{try}\mathrm{parametric}\mathrm{method}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^4-{\mathrm{x}}^2+\mathrm{a}}\mathrm{with}\mathrm{a}>\frac{1}{4}\mathrm{we}\mathrm{have}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4-{\mathrm{x}}^2+\mathrm{a})}^2}$$$$\mathrm{I}=-{\mathrm{f}}^{{}^{\prime }}\left(1\right)\mathrm{let}\mathrm{explicit}\mathrm{f}\left(\mathrm{a}\right)$$$$\phi \left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}^4-{\mathrm{z}}^2+\mathrm{a}}\mathrm{poles}\mathrm{of}\phi ?$$$${\mathrm{u}}^2-\mathrm{u}+\mathrm{a}=0(\mathrm{u}={\mathrm{z}}^2)\to \mathrm{\Delta }=1-4\mathrm{a}<0\Rightarrow {\mathrm{u}}_1=\frac{1+\mathrm{i}\sqrt{4\mathrm{a}-1}}{2}$$$${\mathrm{u}}_2=\frac{1-\mathrm{i}\sqrt{4\mathrm{a}-1}}{2}\mathrm{we}\mathrm{have}\mid {\mathrm{u}}_1\mid =\frac{1}{2}\sqrt{1+4\mathrm{a}-1}=\sqrt{\mathrm{a}}\Rightarrow$$$${\mathrm{u}}_1=\sqrt{\mathrm{a}}{\mathrm{e}}^{\mathrm{iar}4\tan \sqrt{4\mathrm{a}-1}}\mathrm{and}{\mathrm{u}}_2=\sqrt{\mathrm{a}}{\mathrm{e}}^{-\mathrm{iarctan}\left(\sqrt{4\mathrm{a}-1}\right)}\Rightarrow$$$${\mathrm{z}}^4-{\mathrm{z}}^2+\mathrm{a}=({\mathrm{z}}^2-\sqrt{\mathrm{a}}{\mathrm{e}}^{\mathrm{iarctan}(\sqrt{4\mathrm{a}}-1)})({\mathrm{z}}^2-\sqrt{\mathrm{a}}{\mathrm{e}}^{-\mathrm{iarftan}\left(\sqrt{4\mathrm{a}-1}\right)})\Rightarrow$$$$\phi \left(\mathrm{z}\right)=\frac{1}{(\mathrm{z}-\sqrt{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)})(\mathrm{z}+\sqrt{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)})(\mathrm{z}-\sqrt{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)})(\mathrm{z}+\sqrt{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{\{\mathrm{a}-1}\right)})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\sqrt{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan (\sqrt{4\mathrm{a}}-1})+\mathrm{Res}(\phi ,-\sqrt{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)}\}$$$$\mathrm{Res}(\phi ,\left({}^4\sqrt{\mathrm{a}}\right){\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)})$$$$=\frac{1}{2\left({}^4\sqrt{\mathrm{a}}\right){\mathrm{e}}^{\frac{\mathrm{i}}{2}\mathrm{ar}4\tan \left(\sqrt{4\mathrm{a}-1}\right)}\times \sqrt{\mathrm{a}}(2\mathrm{i}\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}$$$$=\frac{{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)}}{4\mathrm{i}{\mathrm{a}}^{\frac{3}{4}}\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}$$$$\mathrm{Res}(\phi ,-\left({}^4\sqrt{\mathrm{a}}\right){\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)})$$$$=\frac{1}{-2\left({}^4\sqrt{\mathrm{a}}\right){\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)}\left(\sqrt{\mathrm{a}}\right)(-2\mathrm{i}\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}$$$$=\frac{{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)}}{4\mathrm{i}{\mathrm{a}}^{\frac{3}{4}}\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{{\mathrm{a}}^{-\frac{3}{4}}}{4\mathrm{i}}\times \frac{{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)}}{\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}+\frac{{\mathrm{a}}^{-\frac{3}{4}}}{4\mathrm{i}}\times \frac{{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{4\mathrm{a}-1}\right)}}{\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}\}$$$$=\frac{\pi }{2{\mathrm{a}}^{\frac{3}{4}}\sin (\arctan \left(\sqrt{4\mathrm{a}-1}\right)}\times 2\cos (\arctan \left(\sqrt{4\mathrm{a}-1}\right)$$$$....\mathrm{be}\mathrm{continued}....$$

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