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Show that $$\int_{0}^{4e}x^2{\ln{x}\ln{(4e-x)}\over \sqrt{x(4e-x)}}\mathrm dx=-e^2\pi(\pi-1)(\pi+1)\tag1$$

My work. Let $x=4e\sin^2{t}$ then $\mathrm dx=8e\sin{t}\cos{t}=4e\sin{(2t)}\mathrm dt$. After a long simplification we got $$32e^2\int\sin^4{t}\ln(4e\sin^2{t})\ln(4e\cos^2{t})\mathrm dt\tag2$$ $(2)$ probably a long IBP.

Robert Z
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gymbvghjkgkjkhgfkl
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  • It looks a bit like $\int x^2f’(x)f(x)g’(x)g(x)\text dx$, but I’m not seeing an easy route to take just yet. – abiessu Dec 19 '17 at 12:52
  • $(1)$ instantly boils down (by removing the useless clutter) to the evaluation of $$ \int_{0}^{1}\frac{x^2 \log(x)\log(1-x)}{\sqrt{x(1-x)}},dx $$ and you should be well-aware that similar integrals can be evaluated by computing derivatives of the Beta function, since it is the seventh or eighth time I provide almost the same answer to one of your questions. Time to ask, read and learn, rather than ask & read only. – Jack D'Aurizio Dec 19 '17 at 21:32
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    I have always been glad to provide solutions to interesting problems, but it is not the first time I raise the same point, i.e. what is the point of asking, if you are not interested in learning something from the answers you receive? This question is just an abstract duplicate of this previous question of yours, and my wonder about why are you doing this? is genuine. – Jack D'Aurizio Dec 19 '17 at 21:56

2 Answers2

3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Note that $\ds{\mc{I}\pars{\mu,\nu} \equiv \int_{0}^{1}x^{\mu}\pars{1 - x}^{\nu} \,\dd x\quad}$ is equal to $\ds{\quad{\Gamma\pars{\mu + 1}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + \nu + 2}}}$ such that

\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} $$ \vphantom{\Huge A^{A^{A}}}\mbox{} $$ \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}\tag{2} \end{equation}

\eqref{1} and \eqref{2} lead to

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

Felix Marin
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Hint. Let $x=2e(1-\cos(t))$, then the given integral becomes $$I=4e^2\int_{0}^{\pi}(1-\cos(t))^2\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt.$$ Now by using the result of your previous question in Show that $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$ , and symmetry, we get $$\begin{align} I&=4e^2\int_{0}^{\pi}\left(\frac{3}{2}-2\cos(t)+\frac{\cos(2t)}{2}\right)\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt\\ &=e^2\pi(6−\pi^2)+0+2e^2\int_{0}^{\pi}\cos(2t)\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt.\end{align}$$ Can you take it from here? For the last integral use the same Fourier series given in Show that $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$ and the fact that $$2\cos(2t)\cos(kt)=\cos((k-2)t)+\cos((k+2)t).$$

Robert Z
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