Cálculo Pregunta: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$

Question

Alguien me puede ayudar a encontrar $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$? Cualquier ayuda se agradece. Gracias de antemano.

Answer 1

Reescribir $$ \begin{align} \int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\ dx. \end{align} $$ Deje $\,u=\cos x$,$\,du=-\sin x\,dx$. Para$\,0 < x < \frac{\pi}{2}$,$\,0 < u < 1$. Ahora, la integral resulta ser $$ \begin{align} \int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\,dx&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sqrt{1-\cos^2 x})\,dx\\ &=-\frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du.\tag1\\ \end{align} $$ A continuación, utilice la serie de Maclaurin para el logaritmo natural: $$ \ln(1-u^2)=-\sum_{n=1}^\infty \frac{u^{2n}}{n}.\tag2\\ $$ Sustituto $\,(2)$$\,(1)$, el rendimiento de $$ \begin{align} \frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du&=-\frac{1}{2}\int_0^1\sum_{n=1}^\infty \frac{u^{2n}}{un}\,du\\ &=-\frac{1}{2}\sum_{n=1}^\infty\int_0^1 \frac{u^{2n-1}}{n}\,du\\ &=-\frac{1}{4}\sum_{n=1}^\infty \left.\frac{u^{2n}}{n^2}\right|_{u=0}^1\\ &=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}.\tag3 \end{align} $$ La serie infinita en $(3)$ se define como Riemann zeta función de $\,\zeta (2)=\dfrac{\pi^2}{6}$. Por lo tanto, $$ \begin{align} \int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}\\ &=-\frac{1}{4}\cdot \frac{\pi^2}{6}\\ &=\large\color{blue}{-\frac{\pi^2}{24}}.\\ \end{align} $$

Answer 2

Aquí está una línea de prueba usando la función beta

$$\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx=\lim_{a\rightarrow 1,b\rightarrow 1^+}\frac{1}{2}\frac{\partial B}{\partial a}\left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)=-\frac{\pi^2}{24}$$

Q. E. D.

Answer 3

$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)log(\sin(x))}{\cos(x)}dx$$

Deje $t=\cos(x), \;\ \sin(x)=\sqrt{1-t^{2}}, \;\ dx=\frac{-1}{\sqrt{1-t^{2}}}dt$

$$1/2\int_{0}^{1}\frac{log(1-t^{2})}{t}dt$$

Esta integral se puede hacer uso de la serie de $$log(1-t^{2})=-\sum_{k=1}^{\infty}\frac{t^{2k}}{k}$$, y es bastante famoso.

Pero, se evalúa a $$\frac{-\pi^{2}}{24}$$

Answer 4

Subst. $t=\sin{x}$, entonces la integral dada se convierte en \begin{align*} \int_0^1 \, \frac{t\, \log{t}}{1-t^2} dt &= \int_{0}^{1} \, \log{t} \, \sum_{k\ge 0} t^{2k+1} \, dt\\ &= \sum_{k\ge 0} \int_{0}^{1} \, \left(\log{t}\right)\, t^{2k+1}\, dt\\ &= \sum_{k\ge 0} -\frac{1}{4 \, {\left(k^{2} + 2 \, k + 1\right)}}\\ &= -\frac{1}{4} \zeta{(2)}\\ &= -\frac{\pi^2}{24} \end{align*}

En general, podemos tener \begin{align*} \int_{0}^{\pi/2} \, \left(\log{\sin{x}}\right)^n\, \tan{x}\, dx = (-1)^n\, \frac{ n!\, \zeta(n + 1)}{2^{n + 1}} \end{align*}

Answer 5

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\overbrace{\color{#00f}{\large\int_{0}^{\pi/2}\tan\pars{x}\ln\pars{\sin\pars{x}}\,\dd x}} ^{\ds{t\ \equiv \sin\pars{x}}}\ =\ \overbrace{\int_{0}^{1}{t\ln\pars{t} \over 1 - t^{2}}\,\dd t} ^{\ds{t\ \equiv \expo{-\xi}}}\ =\ -\int_{0}^{\infty}{\xi\expo{-2\xi} \over 1 - \expo{-2\xi}}\,\dd\xi \\[3mm]&= -\,{1 \over 4}\int_{0}^{\infty}{\xi\expo{-\xi} \over 1 - \expo{-\xi}}\,\dd\xi =\,{1 \over 4}\int_{0}^{\infty}\ln\pars{1 - \expo{-\xi}}\,\dd\xi ={1 \over 4}\int_{0}^{\infty} \pars{-\sum_{n = 1}^{\infty}{\expo{-n\xi} \over n}}\,\dd\xi \\[3mm]&=-\,{1 \over 4}\sum_{n = 1}^{\infty}{1 \over n}\ \underbrace{\int_{0}^{\infty}\expo{-n\xi}\,\dd\xi}_{\ds{=\ {1 \over n}}} =-\,{1 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{=\ {\pi^{2} \over 6}}} = \color{#00f}{\Large -\,{\pi^{2} \over 24}} \end{align}