Integral $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx$, donde $0<c<1$

Question

Estoy intentando evaluar la siguiente integral:

$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx$$

donde $0

No se me ocurre realmente una forma de encontrarla, así que por favor dame una pista.

Answer 1

Expresar el integrando como una serie:

$$\begin{align}I(c) &= \int_{-\pi/2}^{\pi/2} dx \, \log{(1+c \sin{x})} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} c^k \int_{-\pi/2}^{\pi/2} dx \, \sin^k{x} \\ &= - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{c^{2 k}}{k} \frac1{2^{2 k}} \binom{2 k}{k} \end{align} $$

$$I'(c) = -\pi \sum_{k=1}^{\infty} \frac1{2^{2 k}} \binom{2 k}{k} c^{2 k-1} = -\frac{\pi}{c} \sum_{k=1}^{\infty} \frac1{2^{2 k}} \binom{2 k}{k} c^{2 k}= -\frac{\pi}{c} \left (\frac1{\sqrt{1-c^2}}-1\right )$$

$$\implies I(c) = -\pi \int dc \frac{1}{c} \left (\frac1{\sqrt{1-c^2}}-1\right ) = \pi \int dt \, (\sec{t}-\tan{t}) = \pi \log{(1+\sin{t})} + K$$

o

$$I(c) = K +\pi \log{\left ( 1+\sqrt{1-c^2} \right )} $$

$$I(0) = 0 \implies K=-\pi \log{2}$$

Por lo tanto

$$I(c) = \pi \log{\left ( \frac{1+\sqrt{1-c^2}}{2}\right )} $$

Answer 2

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\begin{align}&\color{#66f}{\large% \int_{-\pi/2}^{\pi/2}\ln\pars{1 + c\sin\pars{x}}\,\dd x} =\int_{-\pi/2}^{\pi/2}\ \overbrace{% \int_{0}^{1}\frac{c\sin\pars{x}\,\dd t}{1 + c\sin\pars{x}t}} ^{\dsc{\ln\pars{1 + c\sin\pars{x}}}}\ \,\dd x \\[5mm]&=\int_{0}^{1} \int_{-\pi/2}^{\pi/2} \frac{\bracks{1 + ct\sin\pars{x}} - 1\,\dd x}{1 + ct\sin\pars{x}}\,\frac{\dd t}{t} =\int_{0}^{1}\bracks{\pi -\ \overbrace{\int_{-\pi/2}^{\pi/2} \frac{\,\dd x}{1 + ct\sin\pars{x}}} ^{\ds{\dsc{\xi} = \dsc{\tan\pars{\frac{x}{2}}}}}}\,\frac{\dd t}{t} \\[5mm]&=\int_{0}^{1}\bracks{\pi - \int_{1}^{1} \frac{1}{1 + ct\pars{2\xi}/\pars{1 + \xi^{2}}}\, \frac{2\,\dd\xi}{1 + \xi^{2}}}\,\frac{\dd t}{t} =2\int_{0}^{1} \bracks{\frac{\pi}{2} - \int_{-1}^{1}\frac{\dd\xi}{\xi^{2} + 2ct\xi + 1}}\, \frac{\dd t}{t} \\[5mm]&=2\int_{0}^{1}\bracks{% {\pi \over 2} -\int_{-1 + ct}^{1 + ct}\frac{\dd\xi}{\xi^{2} + 1 - c^{2}t^{2}}}\,\frac{\dd t}{t} \\[5mm]&=2\int_{0}^{1}\braces{{\pi \over 2} -\frac{1}{\root{1 - c^{2}t^{2}}}\ \overbrace{\bracks{% \arctan\pars{\frac{1 + ct}{\root{1 - c^{2}t^{2}}}} -\arctan\pars{\frac{-1 + ct}{\root{1 - c^{2}t^{2}}}}}}^{\ds{=\dsc{\frac{\pi}{2}}}}} \,\frac{\dd t}{t} \\[5mm]&=\pi\int_{0}^{1}\bracks{1 -\frac{1}{\root{1 - c^{2}t^{2}}}}\,\frac{\dd t}{t} =\color{#66f}{\large\pi\ln\pars{\frac{1 + \root{1 - c^{2}}}{2}}} \end{align>

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Note that \begin{align>&\overbrace{% \int\frac{1}{\root{1 - c^{2}t^{2}}}\,\frac{\dd t}{t}}^{\ds{\dsc{ct}=\dsc{\cos\pars{\theta}}}}\ =\ \int\frac{1}{\sin\pars{\theta}}\,\frac{-\sin\pars{\theta}\,\dd\theta/c}{\cos\pars{\theta}/c} =-\int\sec\pars{\theta}\,\dd\theta \\[5mm]&=-\ln\pars{\sec\pars{\theta} + \tan\pars{\theta}} =-\ln\pars{\frac{1 + \root{1 - \cos^{2}\pars{\theta}}}{\cos\pars{\theta}}} \\[5mm]&=-\ln\pars{\frac{1 + \root{1 - c^{2}t^{2}}}{ct}} + \mbox{a constant} \end{align> such that \begin{align> &\pi\int_{0}^{1}\bracks{1 -\frac{1}{\root{1 - c^{2}t^{2}}}}\,\frac{\dd t}{t} =\pi\bracks{\ln\pars{t} + \ln\pars{\frac{1 + \root{1 - c^{2}t^{2}}}{ct}}}_{0}^{1} \\[5mm]&=\left.\pi\ln\pars{\frac{1 + \root{1 - c^{2}t^{2}}}{c}} \right\vert_{\, 0}^{\, 1}=\pi\ln\pars{\frac{1 + \root{1 - c^{2}}}{2}} \end{align>

Answer 3

Una idea instantánea:

Introduce la función $$f(c) =\int_{-\pi/2}^{\pi/2}{\log(1+c\sin(x))dx}$$ Que satisface el valor inicial $f(0) =0$

El siguiente paso es diferenciar bajo el signo de la integral y obtener una forma cerrada de la derivada.

Answer 4

ESTA ES SOLO UNA RESPUESTA PARCIAL

Primero aplicamos DUIS $$I'(a)=\int_{-\pi/2}^{\pi/2} \frac{\sin(x)}{a\sin(x)+1}$$ Esto se puede integrar con una sustitución de Weierstrass, que dejaré para que verifiques: $$\large\left.I'(a)=\frac{x-\frac{2\tan^{-1}\left(\frac{\tan(x/2)+y}{\sqrt{1-y^2}}\right) }{\sqrt{1-y^2}}}{y}\right|_{-\pi/2}^{\pi/2}$$

Después de sustituir los límites obtenemos: $$\large I'(a)=\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}+\frac{\pi/2+\frac{2\tan^{-1}\left(\frac{-1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}=I_1+I_2$$

Ahora tenemos que integrar ese enorme monstruo.

$$\large I_1=\int\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}$$ $$\large=\int\cos\theta\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\cos\theta}}{\sin\theta}(y=\sin\theta)$$

$$\large=\int\cot\theta\left({\pi/2-\frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\cos\theta}}\right)$$

$$\large=\int\frac{\pi\cot\theta}2 +\int \frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\sin\theta}$$

$$\large=\frac \pi 2 \log \sin\theta +\int \frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\sin\theta}$$

Lo cual no es expresable en términos de funciones elementales.

Answer 5

Permita que $c=\sin a $\begin{align} \int_{-\pi/2}^{\pi/2} \ln(1+\sin a\sin x)dx =& \int_{-\pi/2}^{\pi/2}\int_0^a \frac{\cos t\sin x}{1+\sin t\sin x}dt\ dx \\ =& -\pi\int_0^a \tan\frac t2\ dt=2\pi\ln \left(\cos \frac a2\right)\\ =&\ \pi\ln\frac{1+\sqrt{1-c^2}}2 \end{align}