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Mi trabajo. Que $$\int_{0}^{4e}x^2{\ln{x}\ln{(4e-x)}\over \sqrt{x(4e-x)}}\mathrm dx=-e^2\pi(\pi-1)(\pi+1)\tag1$ y $x=4e\sin^2{t}$. Después de una larga simplificación tenemos $\mathrm dx=8e\sin{t}\cos{t}=4e\sin{(2t)}\mathrm dt$$ $$32e^2\int\sin^4{t}\ln(4e\sin^2{t})\ln(4e\cos^2{t})\mathrm dt\tag2$ probablemente un IBP largo.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{4\expo{}}x^{2}\,{\ln\pars{x}\ln\pars{4\expo{} - x} \over \root{x\pars{4\expo{} - x}}}\,\dd x}} \,\,\,\stackrel{x/\pars{4\expo{}}\ \mapsto\ x}{=}\,\,\, \int_{0}^{1}\pars{4\expo{}}^{2}x^{2}\, {\ln\pars{4\expo{}x}\ln\pars{4\expo{} - 4\expo{}x} \over \root{4\expo{}x\pars{4\expo{} - 4\expo{}x}}}\,4\expo{}\,\dd x \\[5mm] = &\ 16\expo{}^{2}\int_{0}^{1}x^{2}\,{\bracks{\ln\pars{4\expo{}} + \ln\pars{x}} \bracks{\ln\pars{4\expo{}} + \ln\pars{1 - x}} \over \root{x\pars{1 - x}}}\,\dd x \\[5mm] = &\ 16\expo{}^{2}\left[% \ln^{2}\pars{4e}\!\!\int_{0}^{1}\!\!x^{3/2}\pars{1 - x}^{-1/2}\,\dd x + \ln\pars{4e}\!\!\int_{0}^{1}\!\!{x^{2}\ln\pars{x} \over \root{x\pars{1 - x}}} \,\dd x + \ln\pars{4e}\!\!\int_{0}^{1}\!\!{x^{2}\ln\pars{1 - x} \over \root{x\pars{1 - x}}}\,\dd x\right. \\[2mm] & \phantom{AAAAA}+ \left.\int_{0}^{1}\!\!{x^{2}\ln\pars{x}\ln\pars{1 - x} \over \root{x\pars{1 - x}}}\,\dd x\right] \end{align}

Tenga en cuenta que $\ds{\mc{I}\pars{\mu\nu} \equiv \int_{0}^{1}x^{\mu}\pars{1 - x}^{\nu} \,\dd x\quad}$ es igual a $\ds{\quad{\Gamma\pars{\mu + 1}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + \nu + 2}}}$ tal que

\begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{4\expo{}}x^{2}\,{\ln\pars{x}\ln\pars{4\expo{} - x} \over \root{x\pars{4\expo{} - x}}}\,\dd x}} \\[5mm] = &\ 16\expo{}^{2}\left[% \ln^{2}\pars{4\expo{}}\,\mc{I}\pars{{3 \over 2},-\,{1 \over 2}} + \ln\pars{4\expo{}}\,\left.\partiald{\mc{I}\pars{\mu,-1/2}}{\mu} \right\vert_{\ \mu\ =\ 3/2} + \ln\pars{4\expo{}}\,\left.\partiald{\mc{I}\pars{3/2,\nu}}{\nu} \right\vert_{\ \nu\ =\ -1/2}\right. \\[2mm] & \phantom{AAAAA}+ \left.{\left.{\partial^{2}\mc{I}\pars{\mu,\nu} \over \partial\mu\,\partial\nu}\right\vert_{\ \mu\ =\ 3/2\,,\ \nu\ =\ -1/2}}\right] \label{1}\tag{1} \end{align} $$ \vphantom{\Enorme A^{^{A}}}\mbox{} $$ \begin{equation} \mbox{with}\quad\left\{\begin{array}{rcl} \ds{\mc{I}\pars{{3 \over 2},-\,{1 \over 2}}} & \ds{=} & \ds{\phantom{-}{3\pi \over 8}} \\[5mm] \ds{\left.\partiald{\mc{I}\pars{\mu,-1/2}}{\mu} \right\vert_{\ \mu\ =\ 3/2}} & \ds{=} & \ds{\phantom{-}{7\pi \over 16} - {3\pi\ln\pars{2} \over 4}} \\[5mm] \ds{\left.\partiald{\mc{I}\pars{3/2,\nu}}{\nu} \right\vert_{\ \nu\ =\ -1/2}} & \ds{=} & \ds{-\,{9\pi \over 16} - {3\pi\ln\pars{2} \over 4}} \\[5mm] \ds{\left.{\partial^{2}\mc{I}\pars{\mu,\nu} \over \partial\mu\,\partial\nu}\right\vert_{\ \mu\ =\ \nu\ =\ -1/2}} & \ds{=} & \ds{-\,{3\pi \over 16} - {\pi^{3} \over 16} + {\pi\ln\pars{2} \over 4} + {3\pi\ln^{2}\pars{2} \over 2}} \end{array}\right. \label{2}\etiqueta{2} \end{equation}

\eqref{1} y \eqref{2} conducir a

$$ \bbx{\int_{0}^{4\expo{}}x^{2}\,{\ln\pars{x}\ln\pars{4\expo{} - x} \over \raíz{x\pars{4\expo{} - x}}}\,\dd x = -\expo{}^{2}\pi\pars{\pi - 1}\pars{\pi + 1}} $$

Answer 2

Sugerencia. Deje $x=2e(1-\cos(t))$, entonces la integral dada se convierte en $$I=4e^2\int_{0}^{\pi}(1-\cos(t))^2\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt.$$ Ahora con el resultado de tu pregunta anterior en Mostrar que $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$ , y la simetría, obtenemos $$\begin{align} I&=4e^2\int_{0}^{\pi}\left(\frac{3}{2}-2\cos(t)+\frac{\cos(2t)}{2}\right)\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt\\ &=e^2\pi(6−\pi^2)+0+2e^2\int_{0}^{\pi}\cos(2t)\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt.\end{align}$$ Se puede tomar desde aquí? Para el último uso integral de la misma serie de Fourier dada en Mostrar que $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$ y el hecho de que $$2\cos(2t)\cos(kt)=\cos((k-2)t)+\cos((k+2)t).$$