Integral

Question

Inspirado en este tema , ¿puedo probar fácilmente el resultado a continuación?

ps

La antiderivada elemental existe, pero parece masoquista si se quiere calcularla.

Answer 1

Como escribió, un CAS puede encontrar la antiderivada$$f(x)=\int x \cos(nx)\log (\tan (x))\,dx$$ and it is quite ugly (involving hypergeometric series with arguments $ e ^ ​​{4ix} $).

Por lo tanto, está claro que, si$n$ es un múltiplo de$4$, habrá muchas simplificaciones para$$g(n)=\int_0^{\frac\pi 2} x \cos(nx)\log (\tan (x))\,dx$ $ como se muestra a continuación para las primeras. $ \ Left (\begin{array}{cc} n & g(n) \\ 4 & \frac{3}{4} \\ 8 & \frac{13}{36} \\ 12 & \frac{211}{900} \\ 16 & \frac{7613}{44100} \\ 20 & \frac{270407}{1984500} \end {array} \ right)$$ What is also interesting is that for other even values of $ {\ pi ^ 2} $, the result is just a multiple of $ 4n $.

Por supuesto, para estos casos específicos y en particular el que has publicado, como comentó @tired, la integración por partes conduce a un resultado "bastante sencillo".

Answer 2

Unesdoc.unesco.org unesdoc.unesco.org

Debido a$\displaystyle \int x \cos(8x) \ln(\tan x) dx =$ y$\displaystyle \frac{1}{576}(9\ln(\cos x)+72x\sin(8x)\ln(\tan x)+9\cos(8x)\ln(\tan x)-$

obtenemos $\displaystyle 9\ln(\sin x)-144x\sin(2x)-48x\sin(6x)-90\cos(2x)-14\cos(6x)) + C$ .

Answer 3

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

$\ds{\int_{0}^{\pi/2}x\cos\pars{8x}\ln\pars{\tan\pars{x}}\,\dd x = {13 \más de 36}:\ {\large ?}}$.

\begin{equation}\bbx{% \mbox{Note that}\ \int_{0}^{\pi/2}\!\!\!\!\!\!\!\!x\cos\pars{8x}\ln\pars{\tan\pars{x}}\,\dd x = \left.\partiald{}{\nu}\bbox[#ffd]{\ds{\int_{0}^{\pi/2}\!\!\!\!\!\!\! \sin\pars{\nu x}\ln\pars{\tan\pars{x}}\,\dd x}}\,\right\vert_{\,\,\nu\ =\ 8}} \label{1}\tag{1} \end{equation} \begin{align} &\left.\bbox[#ffd,10px]{\ds{\int_{0}^{\pi/2} \sin\pars{\nu x}\ln\pars{\tan\pars{x}}\,\dd x}} \right\vert_{\ \nu\ \in\ \mathbb{R}_{\ >\ 0}} = \Im\int_{0}^{\pi/2}\expo{\ic\nu x}\ln\pars{\tan\pars{x}}\,\dd x \\[5mm] = &\ \left.\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}z^{\nu}\ln\pars{{1 - z^{2} \over 1 + z^{2}}\,\ic}{\dd z \over \ic z}\, \right\vert_{\ z\ =\ \exp\pars{\ic x}}\qquad\pars{~\substack{\ds{z^{\nu}\ \mbox{and}\ \ln\ \mbox{are defined as}}\\[1mm] \ds{Principal\ Branches}}~} \\[5mm] = &\ \left.-\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}z^{\nu - 1} \ln\pars{{1 - z^{2} \over 1 + z^{2}}\,\ic}\dd z\, \right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,& \Re\int_{1}^{0}\pars{\ic y}^{\nu - 1} \ln\pars{{1 + y^{2} \over 1 - y^{2}}\,\ic}\ic\,\dd y + \Re\int_{0}^{1}x^{\nu - 1} \ln\pars{{1 - x^{2} \over 1 + x^{2}}\,\ic}\,\dd x \end{align}

La integración se realizó en un cuarto de círculo en el primer cuadrante. Las 'contribuciones' de sangría caminos 'alrededor de' $\ds{z = \ic}$ $\ds{z = 1}$ fueron omitidos debido a que se desvanece como $\ds{\epsilon \to 0^{+}}$.

A continuación, \begin{align} &\left.\bbox[#ffd,10px]{\ds{\int_{0}^{\pi/2} \sin\pars{\nu x}\ln\pars{\tan\pars{x}}\,\dd x}} \right\vert_{\ \nu\ \in\ \mathbb{R}_{\ >\ 0}} \\[5mm] = &\ -\Re\int_{0}^{1}\expo{\ic\nu\pi/2}y^{\nu - 1} \bracks{\ln\pars{1 + y^{2} \over 1 - y^{2}} + \ic\,{\pi \over 2}}\,\dd y + \Re\int_{0}^{1}x^{\nu - 1} \bracks{\ln\pars{1 - x^{2} \over 1 + x^{2}} + \ic\,{\pi \over 2}}\,\dd x \\[5mm] = &\ -\cos\pars{\nu\pi \over 2}\int_{0}^{1}y^{\nu - 1} \ln\pars{1 + y^{2} \over 1 - y^{2}}\,\dd y + {\pi \over 2}\sin\pars{\nu\pi \over 2}\int_{0}^{1}y^{\nu - 1}\,\dd y + \int_{0}^{1}x^{\nu - 1}\ln\pars{1 - x^{2} \over 1 + x^{2}}\,\dd x \\[5mm] & = 2\cos^{2}\pars{\nu\pi \over 4} \int_{0}^{1}x^{\nu - 1}\ln\pars{1 - x^{2} \over 1 + x^{2}}\,\dd x + {\pi\sin\pars{\nu\pi/2} \over 2\nu} \end{align}

\eqref{1} se reduce a:

\begin{align} &\int_{0}^{\pi/2}x\cos\pars{8x}\ln\pars{\tan\pars{x}}\,\dd x = 2\int_{0}^{1}x^{7}\ln\pars{x}\ln\pars{1 - x^{2} \over 1 + x^{2}}\,\dd x + {\pi^{2} \over 32} \\[5mm] = &\ 4\int_{0}^{1}x^{7}\ln\pars{x}\ln\pars{1 - x^{2}}\,\dd x - 2\int_{0}^{1}x^{7}\ln\pars{x}\ln\pars{1 - x^{4}}\,\dd x + {\pi^{2} \over 32} \\[5mm] = &\ \underbrace{\int_{0}^{1}x^{3}\ln\pars{x}\ln\pars{1 - x}\,\dd x} _{\ds{{35 \over 72} - {\pi^{2} \over 24}}}\ -\ {1 \over 8}\ \underbrace{\int_{0}^{1}x\ln\pars{x}\ln\pars{1 - x}\,\dd x} _{\ds{1 - {\pi^{2} \over 12}}}\ +\ {\pi^{2} \over 32} = \bbx{13 \over 36} \approx 0.3611 \end{align}