Cómo mostrar que$\int_{0}^{\infty}{1\over x}\ln\left({1\over x}\right)\sin(2x)\sin^2(x)\mathrm dx={\pi \gamma \over 8}?$

Question

Dado que:

Donde$\gamma$ es la constante de Euler-Masheroni

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Answer 1

El problema se reduce a calcular$$ J(k) = \int_{0}^{+\infty}\frac{\log x}{x}\sin(kx)\,dx \tag{1}$ $ y por la transformación de Laplace $$ \mathcal{L}^{-1}\left(\frac{\log x}{x}\right) = -\gamma-\log(s),\qquad \mathcal{L}\left(\sin(kx)\right)= \frac{k}{k^2+s^2}\tag{2} $ $ por lo tanto es fácil comprobar que para cualquier$k>0$:$$ J(k)=\color{red}{-\frac{\pi}{2}\left(\gamma+\log k\right)}.\tag{3} $ $

Answer 2

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$\ds{\int_{0}^{\infty}{1\over x} \ln\pars{1\over x}\sin\pars{2x}\sin^{2}\pars{x}\,\dd x = {\pi\gamma \más de 8}:\ {\large ?}}$.

\begin{align} &\int_{0}^{\infty}{1\over x} \ln\pars{1\over x}\sin\pars{2x}\sin^{2}\pars{x}\,\dd x = -\int_{0}^{\infty} \ln\pars{x}\,{1\over x}\,\sin\pars{2x}\,{1 - \cos\pars{2x} \over 2}\,\dd x \\[5mm] = &\ \int_{0}^{\infty}\ln\pars{x} \bracks{{\sin\pars{4x} \over 4x} - {\sin\pars{2x} \over 2x}}\,\dd x = \int_{0}^{\infty}\ln\pars{x}{\sin\pars{4x} \over 4x}\,\dd x - \int_{0}^{\infty}\ln\pars{x}{\sin\pars{2x} \over 2x}\,\dd x \\[5mm] = &\ -\,{1 \over 4}\int_{0}^{\infty}\ln\pars{x}{\sin\pars{x} \over x}\,\dd x = -\,{1 \over 4}\int_{0}^{\infty}\ln\pars{x}\sin\pars{x} \int_{0}^{\infty}\expo{-xt}\,\dd t\,\dd x \\[5mm] = &\ -\,{1 \over 4}\,\Im\int_{0}^{\infty} \int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\dd x\,\dd t \\[5mm] = &\ -\,{1 \over 4}\,\Im\int_{0}^{\infty}{1 \over t - \ic} \int_{0}^{\pars{t - \ic}\infty}\ln\pars{x \over t - \ic}\expo{-x}\dd x\,\dd t \qquad\pars{~\ln:\ Principal\ Branch~} \\[5mm] = &\ -\,{1 \over 4}\,\Im\int_{0}^{\infty}{1 \over t - \ic} \int_{0}^{\infty}\ln\pars{x \over t - \ic}\expo{-x}\dd x\,\dd t\qquad \pars{\substack{\mbox{I omitted an integral along}\\[1mm] \mbox{an arc, in the complex plane,}\\[1mm] \mbox{ of radius}\ R \to \infty.\\[2mm] \mbox{Such integral vanishes out as}\ R\to\ \infty}} \\[5mm] = &\ -\,{1 \over 4}\ \overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\ \overbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}^{\ds{=\ -\gamma}}\ +\ {1 \over 4}\ \overbrace{\quad\Im\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}{\ln\pars{t - \ic} \over t - \ic}\,\dd t\quad} ^{\ds{=\ \Im\pars{-\,{1 \over 2}\,\ln^{2}\pars{-\ic}}}\ =\ {\large 0}}\ =\ \bbx{\pi\gamma \over 8} \end{align}