Cómo mostrar que $\int_{0}^{\pi/4}{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx={\ln(2^2)-\ln^2(2)\over 8}?$

Question

Dado que:

$$\int_{0}^{\pi/4}{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx={\ln(2^2)-\ln^2(2)\over 8}\tag1$$

No estoy seguro de cómo ir sobre para empezar a abordar este problema de la prueba de $(1)$.

Answer 1

Deje $$I=\int{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx\tag1$$ tenemos

$$I=\int \frac{\ln(\cos x)\ln(\sin x) \cos(2x)\:dx}{\sin^2(2x)}$$ Using Integration by Parts taking $u=\ln(\cos x)\ln(\sin x) $ and $v=\frac{\cos (2x)}{\sin^2(2x)}$

Ahora tenemos $$\int v dx=\frac{-1}{2 \sin(2x)}$$

$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}+\int \left(\frac{d}{dx}(\ln(\cos x)\ln(\sin x))\right) \times \frac{-1}{2 \sin(2x)}\:dx$$

así

$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)} -\frac{1}{2}\int \frac{\ln(\sin x)\tan x \:dx}{\sin(2x)}+\frac{1}{2}\int\frac{\ln(\cos x)\cot x \:dx}{\sin(2x)}$$ de ahí

$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}-\frac{1}{4}\int \ln(\sin x)\sec^2 x \: dx+\frac{1}{4}\int \ln(\cos x)\csc^2 x\:dx$$

Pero

$$\int\ln(\sin x)\sec^2 x \:dx=\ln(\sin x)\tan x-x+C$$ y

$$ \int\ln(\cos x)\csc^2 x \:dx=-\ln(\cos x)\cot x-x+C$$ así que, Finalmente,

$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}-\frac{1}{4}\left(\ln(\sin x)\tan x-x\right)+\frac{1}{4}\left(-\ln(\cos x)\cot x-x\right)$$ Lo

$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}-\frac{1}{4}\left(\ln(\sin x)\tan x+\ln(\cos x)\cot x \right)$$

Ahora aplicar los límites de $0$ $\frac{\pi}{4}$y el uso de $$\lim_{x \to 0^+}x \:\ln x=0$$ obtenemos

$$I=\frac{\ln (2^2)-\ln^2 (2)}{8}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \\sin\pars{2x}\tan\pars{2x}}\,\dd x = {\ln\pars{2^{2}}-\ln^{2}\pars{2} \más de 8}:\ {\large ?}}$.

\begin{align} &\int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin\pars{2x}\tan\pars{2x}}\,\dd x = \int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin^{2}\pars{2x}}\,\cos\pars{2x}\,\dd x \\[5mm] \stackrel{2z\ \mapsto\ x}{=}\,\,\, &\ {1 \over 2}\int_{0}^{\pi/2} \ln\pars{\cos\pars{x \over 2}}\ln\pars{\sin\pars{x \over 2}}\, {\cos\pars{x} \over \sin^{2}\pars{x}}\,\dd x \\[5mm] = &\ -\,{1 \over 2}\int_{x\ =\ 0}^{x\ =\ \pi/2} \ln\pars{\cos\pars{x \over 2}}\ln\pars{\sin\pars{x \over 2}}\,\dd\bracks{1 \over \sin\pars{x}} \\[1cm] \stackrel{\mbox{IBP}}{=}\,\,\, &\ -\,{1 \over 8}\,\ln^{2}\pars{2} \\[3mm] &\ + {1 \over 2}\int_{0}^{\pi/2}\!\!\!\!\!\!\!{1 \over \sin\pars{x}}\bracks{% {-\sin\pars{x/2}/2 \over \cos\pars{x/2}}\,\ln\pars{\sin\pars{x \over 2}} + \ln\pars{\cos\pars{x \over 2}}\,{\cos\pars{x/2}/2 \over \sin\pars{x/2}}}\dd x \\[1cm] = &\ -\,{1 \over 8}\,\ln^{2}\pars{2} \\[3mm] + &\ {1 \over 16}\int_{0}^{\pi/2} \ln\pars{\cos^{2}\pars{x \over 2}}{\dd x \over \sin^{2}\pars{x/2}} - {1 \over 16}\int_{0}^{\pi/2}\ln\pars{\sin^{2}\pars{x \over 2}} {\dd x \over \cos^{2}\pars{x/2}} \\[1cm] = &\ -\,{1 \over 8}\,\ln^{2}\pars{2} \\[3mm] + &\ {1 \over 8}\int_{0}^{\pi/2} \ln\pars{1 + \cos\pars{x} \over 2}{\dd x \over 1 - \cos\pars{x}} - {1 \over 8}\int_{0}^{\pi/2}\ln\pars{1 - \cos\pars{x} \over 2} {\dd x \over 1 + \cos\pars{x}} \\[1cm] = &\ -\,{1 \over 8}\,\ln^{2}\pars{2} \\[3mm] + &\ {1 \over 8}\ \underbrace{\int_{0}^{\pi/2} \ln\pars{1 + \cos\pars{x} \over 2}{\dd x \over 1 - \cos\pars{x}}} _{\ds{\equiv\ \,\mc{I}_{1}}}\ -\ {1 \over 8}\ \underbrace{\int_{\pi/2}^{\pi}\ln\pars{1 + \cos\pars{x} \over 2} {\dd x \over 1 - \cos\pars{x}}}_{\ds{\equiv\ \,\mc{I}_{2}}} \\[1cm] = &\ {\mc{I}_{1} - \mc{I}_{2} - \ln^{2}\pars{2} \over 8}\label{1}\tag{1} \end{align}

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$\ds{\Large\mc{I}_{1}\ \mbox{and}\ \,\mc{I}_{2}:\ ?}$ \begin{align} \mc{I}_{1} & \equiv \int_{0}^{\pi/2}\ln\pars{1 + \cos\pars{x} \over 2}{\dd x \over 1 - \cos\pars{x}} \\[5mm] = &\ \left.-\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln\pars{1 + \bracks{z^{2} + 1}/\bracks{2z} \over 2} {1 \over \pars{z^{2} + 1}/\pars{2z} - 1}\,{\dd z \over \ic z}\, \right\vert_{\ z\ \equiv\ \exp\pars{\ic x}} \\[5mm] = &\ \left.-2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln\pars{\bracks{z + 1}^{2} \over 4z} {1 \over \pars{z - 1}^{2}}\,\dd z\, \right\vert_{\ z\ \equiv\ \exp\pars{\ic x}} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, &\ 2\,\Im\int_{1}^{\epsilon}\ln\pars{\bracks{\ic y + 1}^{2} \over 4\ic y} {1 \over \pars{\ic y - 1}^{2}}\,\ic\,\dd y \end{align} El $\ds{\ln}$ función tiene una rama de corte a lo largo de la negativa eje real' con $\ds{-\pi < \,\mrm{arg}\pars{z} < \pi}$. La integración se realizó a lo largo de un cuarto de circunferencia en el primer cuadrante con una sangría 'alrededor $\ds{z = 0}$'. El $\ds{\,\mc{I}_{1}}$ toda contribución proviene de una integral que 'corre' a lo largo del eje vertical y de la mencionada guión. Por simplicidad se omite 'contribuciones' de términos que son reales o se va a cero, como se $\ds{\epsilon \to 0^{+}}$. Entonces, como $\ds{\epsilon \to 0^{+}}$ \begin{align} \mc{I}_{1} & = -2\,\Re\int_{0}^{1}\braces{\ln\pars{1 + y^{2} \over 4y} + \bracks{2\arctan\pars{y} - {\pi \over 2}}\ic} {\exp\pars{2\arctan\pars{y}\ic} \over 1 + y^{2}}\,\dd y \\[5mm] \stackrel{y\ =\ \tan\pars{x}}{=}\,\,\, &\ -2\,\Re\int_{0}^{\pi/4}\braces{-\ln\pars{2\sin\pars{2x}} + \bracks{2x - {\pi \over 2}}\ic}\exp\pars{2\ic x}\,\dd x \\[5mm] = &\ 2\ \overbrace{\int_{0}^{\pi/4}\ln\pars{2\sin\pars{2x}}\cos\pars{2x}\,\dd x} ^{\ds{{1 \over 2}\,\ln\pars{2} - {1 \over 2}}}\ +\ 2\ \overbrace{\int_{0}^{\pi/4}\pars{2x - {\pi \over 2}}\sin\pars{2x}\,\dd x} ^{\ds{{1 \over 2} - {\pi \over 4}}} \\[5mm] &\ \implies \bbx{\mc{I}_{1} = \ln\pars{2} - {\pi \over 2}}.\quad \mbox{Similarly,}\quad\bbx{\mc{I}_{2} = -\ln\pars{2} - {\pi \over 2}}. \end{align}

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La expresión \eqref{1} se convierte en \begin{align} \int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin\pars{2x}\tan\pars{2x}}\,\dd x & = {\bracks{\ln\pars{2} - \pi/2} - \bracks{-\ln\pars{2} - \pi/2} - \ln^{2}\pars{2}\over 8} \\[5mm] & = \bbx{\ln\pars{2^{2}} - \ln^{2}\pars{2}\over 8} \end{align}