Evaluar la integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$

Question

¿Cómo puedo evaluar la siguiente integral?

$$ \int_0^{\pi/2} \log\left(\frac{1 + \cos\left(x\right)}{1 - \cos\left(x\right)}\right)\, \frac{1}{\cos\left(x\right)}\,{\rm d}x\,, \qquad\left\vert\,\,\right\vert \le 1$$

He intentado diferenciar en virtud de la integral con respecto al parámetro de $a$, y también traté de ampliar el registro de plazo en una serie de Taylor y, a continuación, cambiar el orden de integración y totalización. Me encontré con dificultades con ambos enfoques.

Answer 1

Utilice la expansión $|z| < 1$

$$\log{\left ( \frac{1+z}{1-z}\right )} = 2 \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$

Entonces la integral es igual a

$$2 \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \int_0^{\pi/2} dx \, \cos^{2 k}{x}$$

Es sencillo mostrar que

$$\int_0^{\pi/2} dx \, \cos^{2 k}{x} = \frac{1}{2^{2 k}} \binom{2 k}{k} \frac{\pi}{2}$$

Así es el integral $I(a)$

$$I(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \frac{1}{2^{2 k}} \binom{2 k}{k}$$

Podemos evaluar esta cantidad teniendo en cuenta

$$I'(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k}}{2^{2 k}} \binom{2 k}{k} = \pi \left (1-a^2\right)^{-1/2}$$

Integrando con respecto a los $a$ y teniendo en cuenta que $I(0)=0$, encontramos

$$I(a) = \pi \arcsin{a}$$

Answer 2

$$\begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1+ b\cos x}\right)\frac{1}{\cos x}dx &= \int_{0}^{\pi/2} \int_{b}^{a} \frac{1}{1+t \cos x} \ dt \ dx \\ &= \int_{b}^{a} \int_{0}^{\pi/2}\frac{1}{1+t \cos x} \ dx \ dt \end{align}$$

Que $ \displaystyle u = \tan \frac{x}{2}$.

$$\begin{align} &= \int_{b}^{a} \int_{0}^{1} \frac{1}{1+ t \left(\frac{1-u^{2}}{1+u^{2}} \right)} \frac{2}{1+u^{2}} \ du \ dt \\ &= 2 \int_{b}^{a} \int_{0}^{1} \frac{1}{1+t} \frac{1}{1+ \frac{1-t}{1+t} u^{2}} du \ dt \end{align}$$

Que $\displaystyle w = \sqrt{\frac{1-t}{1+t}} u $.

$$ \begin{align} &= 2 \int_{b}^{a} \int_{0}^\sqrt{\frac{1-t}{1+t}} \frac{1}{\sqrt{1-t^{2}}} \frac{1}{1+w^{2}} \ dw \ dt \\ &= 2 \int_{b}^{a} \frac{1}{\sqrt{1-t^{2}}} \arctan \sqrt{\frac{1-t}{1+t}}\ dt \\ &= \int_{b}^{a} \frac{\arccos t}{\sqrt{1-t^{2}}} \ dt \\ &= \frac{1}{2} \Big(\arccos^{2} (b)- \arccos^{2} (a)\Big) \end{align}$$

Entonces

$$ \begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1-a \cos x}\right)\frac{1}{\cos x}dx &= \frac{1}{2} \Big(\arccos^{2} (-a)- \arccos^{2} (a)\Big) \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} - \arcsin (-a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} + \arcsin (a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big(2 \pi \arcsin a\Big) = \pi \arcsin a \end{align}$$

Answer 3

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_0^{\pi/2}\ln\pars{{1 + \cos\pars{x} \más de 1 a\cos\pars{x}}}\, {1 \over \cos\pars{x}}\,\dd x:\ {\large ?}\,.\qquad\qquad\verts{un}\ <\ 1}$.

La idea general es la derivada respecto de $\ds{\quad a\quad}$ a fin de "matar " el "$\ds{\cos\pars{x}}$ plazo " en el denominador:

\begin{align}&\color{#c00000}{\partiald{}{a}\bracks{\int_0^{\pi/2} \ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\,{\dd x \over \cos\pars{x}}}} \\[3mm]&=\int_0^{\pi/2}\bracks{{\cos\pars{x} \over 1 + a\cos\pars{x}} -{-\cos\pars{x} \over 1 - a\cos\pars{x}}}\,{\dd x \over \cos\pars{x}} =2\int_{0}^{\pi/2}{\dd x \over 1 - a^{2}\cos^{2}\pars{x}} \\[3mm]&=2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over \sec^{2}\pars{x} - a^{2}} =2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over \tan^{2}\pars{x} + 1 - a^{2}} =2\int_{0}^{\infty}{\dd x \over x^{2} + 1 - a^{2}} \\[3mm]&={2 \over \root{1 - a^{2}}}\ \overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\ =\ \color{#c00000}{\pi \over \root{1 - a^{2}}} \end{align}

$$\color{#66f}{\large% \int_0^{\pi/2}\ln\pars{{1 + \cos\pars{x} \más de 1 a\cos\pars{x}}}\, {1 \over \cos\pars{x}}\,\dd x} =\int_{0}^{a}{\pi\,\dd t \\raíz{1 - t^{2}}} =\color{#66f}{\large\pi\ \arcsin\pars{un}} $$