Integración

Question

Estoy tratando de evaluar: $$\int_0^{\pi/12} \ln(\tan x)\,dx$ $

Creo que el integral es bastante simple pero estoy teniendo un momento difícil evaluarla. Empecé con el resultado: constante de % $ $$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$G$Dónde está el catalán. Con el cambio de variables $x\rightarrow 3x$ y usando el hecho de que $\tan(3x)=\tan x\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)$, la integral es: $$\int_0^{\pi/12}\ln(\tan x)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}-x\right)\,dx=-\frac{G}{3}$ $ $$\Rightarrow \int_0^{\pi/12}\ln(\tan x)\,dx+\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=-\frac{G}{3}$ $ pero no veo cómo proceder.

Ayuda es apreciada. ¡ Gracias!

Answer 1

Utilizando la serie de Fourier de $\ln(\tan{x})$, \begin{align} &\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\ =&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\ =&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\ =&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\ &\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\\ =&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{G}}\color{#E2062C}{-\frac{1}{2}G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}G}}\\ =&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)G=\large{-\frac{2}{3}G} \end {alinee el}

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Cosas podrían hacerse más claras si escribimos explícitamente los términos de las cantidades. Para las cantidades de rojo ,\begin{align} &-\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{5^2}-\frac{1}{7^2}-\frac{1}{11^2}+\cdots\right)\\ =&-\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2}-\frac{1}{11^2}+\cdots\right)-\frac{1}{2}\left(\frac{1}{3^2}-\frac{1}{9^2}+\frac{1}{15^2}-\cdots\right)\\ =&-\frac{1}{2}G-\frac{1}{2}\cdot\frac{1}{9}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\cdots\right)=-\frac{5}{9}G \end {alinee el}

Answer 2

Primero: $~\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx\qquad(1)$

Prueba:

Que $I=\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$

$\tan(3x)=\tan(x)\tan\big(\dfrac{\pi}{3}-x\big)\tan\big(\dfrac{\pi}{3}+x\big)$

$\displaystyle I= \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_0^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{3}-x\Big)\Big)dx+\int_0^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{3}-x\Big)\Big)dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{3}}^{\tfrac{5\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} \log(\tan(x))dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{5\pi}{12}} \log(\tan(x))dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{2}-x\Big)\Big)dx$

$\tan\Big (\dfrac{\pi}{2}-x\Big)=\dfrac{1}{\tan(x)}$

Entonces: $~\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{12}}\log(\tan(x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-\int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-3\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-3I$

$\displaystyle 2I=\int_0^{\tfrac{\pi}{12}}\log(\tan(x))dx$

Ahora realice el cambio de la variable $u=3x$ en el miembro izquierdo de $(1)$:

$\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\dfrac{2}{3} \int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx$

Desde $~\displaystyle G=-\int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx~$ y $~\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx=-\dfrac{2}{3}G$.

$($ Prueba de encuentran en: constante de representaciones de Catalan, David Bradley, $2001)$.

Answer 3

$\qquad\qquad\qquad\qquad$Hello, there! Cleo just asked me to post this:

$$\int_0^\tfrac\pi{12}\ln(\tan x)~dx=-\dfrac23\cdot\text{Catalan}$$

Answer 4

Reescribiendo las funciones de Lobachevsky en términos de dilogarithms, conseguimos $$\mathcal{I}=-\frac12\Im\left[\operatorname{Li}_2\left(e^{\pi i/6}\right)+\operatorname{Li}_2\left(e^{5\pi i/6}\right)\right]=\frac12\color{blue}{\Im\left[\operatorname{Li}_2\left(e^{-\pi i/6}\right)-\operatorname{Li}_2\left(e^{5\pi i/6}\right)\right]}=-\frac23\mathbf{G},$ $ donde se calculó la expresión azul en esta respuesta utilizando la fórmula de la triplicación de la $\operatorname{Li}_2(z)$.

Explicación: el bloque de edificio básico es el integral $$\int_0^{\pi\alpha}\ln\left(2\sin x\right)dx=-\frac12\Im\operatorname{Li}_2\left(e^{2\pi i\alpha}\right),\qquad \alpha\in\left[0,1/{2}\right].$$ escritura $\ln\tan x=\ln\left(2\sin x\right)-\ln\left(2\sin (\frac{\pi}{2}-x)\right)$ y que $\Im\operatorname{Li}_2(-1)=0$ el uso reduce la integral anterior.

Answer 5

Sugerencia:

Cambio en $u=x+\frac{\pi}{12}$,

$$\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=\int_{0}^{\pi/6}\ln \tan\left(\frac{\pi}{4}+u\right)\,du.$$

El integrando puede ser expresado mediante una serie trigonométrica como:

$$\frac12\ln\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}=\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[(2k-1)x\right]}}{2k-1}$$

$$\implies \ln\tan{\left(\frac{\pi}{4}+u\right)}=2\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[2(2k-1)u\right]}}{2k-1}.$$

A continuación,

$$\begin{align} \int_{0}^{\pi/6}\ln \tan\left(\frac{\pi}{4}+u\right)\,du &=2\int_{0}^{\pi/6}\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[2(2k-1)u\right]}}{2k-1}\,du\\ &=2\sum_{k-1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\int_{0}^{\pi/6}\sin{\left[2(2k-1)u\right]}\,du\\ &=2\sum_{k-1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\cdot\frac{\cos^2{\left(\frac{\pi}{3}(k+1)\right)}}{2k-1}\\ &=2\sum_{k-1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^2}\cos^2{\left(\frac{\pi}{3}(k+1)\right)}.\\ \end {Alinee el} $$

Sumas de este tipo pueden ser fácilmente re-expresadas como componentes de dilogarithms.