$$I=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$$ Donde G es la contante del catalán Mi intento: Integrando por partes, tenemos: $$3I=\int_0^1x^4\ln(1-x)\frac{dx}{\sqrt{1-x^2}}+\int_0^1x^3\sqrt{1-x^2}\frac{dx}{1-x}=S+T$$ $$\frac{x^3}{1-x}=-x^2-x-1+\frac{1}{1-x}$$ Es fácil deducirlo: $$\int_0^1x^2\sqrt{1-x^2}dx=\frac{\pi}{16},\int_0^1x\sqrt{1-x^2}dx=\frac{1}{3},\int_0^1\sqrt{1-x^2}dx=\frac{\pi}{4},\int_0^1\sqrt{1-x^2}\frac{dx}{1-x}=\frac{\pi}{2}+1$$ $$3I=S+\frac{3\pi}{16}+\frac{2}{3}$$ Pero, ¿cómo deducir S?
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$\displaystyle J=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx$
Realizar el cambio de variable $y=\sqrt{1-x}$ ,
$\displaystyle J=4\int_0^1 x^2(1-x^2)^2\sqrt{2-x^2}\ln x$
Realizar el cambio de variable $x=\sqrt{2}y$ ,
$\begin{align} J&=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln\left(\sqrt{2}x\right)dx\\ &=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln x dx+8\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln2 dx\\ &=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln x dx+\\ &8\ln 2\Big[\frac{\mathrm{arcsin}\left( x\right) }{32}-\frac{{{x}^{5}}\cdot {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{2}+\frac{{{x}^{3}}\cdot {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{4}-\frac{x\cdot {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{16}+\frac{x\cdot \sqrt{1-{{x}^{2}}}}{32}\Big]_0^{\tfrac{1}{\sqrt{2}}}\\ &=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln x dx+\dfrac{\pi}{16}\ln 2\\ &=16\left[\left(\frac{\mathrm{arcsin}\left( x\right) }{32}-\frac{{{x}^{5}}{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{2}+\frac{{{x}^{3}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{4}-\frac{x {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{16}+\frac{x\sqrt{1-{{x}^{2}}}}{32}\right)\ln x\right]_0^{\tfrac{1}{\sqrt{2}}}-\\ &16\int_0^{\tfrac{1}{\sqrt{2}}}\left(\frac{\mathrm{arcsin}\left( x\right) }{32}-\frac{{{x}^{5}}{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{2}+\frac{{{x}^{3}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{4}-\frac{x {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{16}+\frac{x\sqrt{1-{{x}^{2}}}}{32}\right)\dfrac{1}{x}dx+\\ &\dfrac{\pi}{16}\ln 2\\ &=-\int_0^{\tfrac{1}{\sqrt{2}}}\left(\frac{\mathrm{arcsin}\left( x\right) }{2 x}-8 {{x}^{4}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+4{{x}^{2}}{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}-{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+\frac{\sqrt{1-{{x}^{2}}}}{2}\right)dx\\ &=-\int_0^{\tfrac{1}{\sqrt{2}}}\left(-8{{x}^{4}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+4{{x}^{2}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}-{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+\frac{\sqrt{1-{{x}^{2}}}}{2}\right)dx-\\ &\dfrac{1}{2}\int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{asin}\left( x\right) }{x}dx\\ &=-\left[\frac{\sqrt{1-{{x}^{2}}} \left( 48 {{x}^{7}}-104 {{x}^{5}}+74{{x}^{3}}-21 x\right) -3\mathrm{arcsin}\left( x\right) }{48}\right]_0^{\tfrac{1}{\sqrt{2}}}-\dfrac{1}{2}\int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx\\ &=\dfrac{\pi}{64}+\dfrac{1}{24}-\dfrac{1}{2}\int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx\\ \end{align}$
En esta última integral realizar el cambio de variable $y=\arcsin x$ ,
$\begin{align} \int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx&=\int_0^{\tfrac{\pi}{4}}x\cot x\, dx\\ &=\Big[x\ln(\sin x)\Big]_0^{\tfrac{\pi}{4}}-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\ &=-\dfrac{\pi}{8}\ln 2-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\ \end{align}$
En esta última integral realizar el cambio de variable $y=\dfrac{\pi}{2}-x$ ,
$\begin{align}\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx&=\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ln(\cos x)\,dx\\ &=\int_{0}^{\tfrac{\pi}{2}}\ln(\cos x)\,dx-\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx\\ &=-\dfrac{\pi}{2}\ln 2-\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx\\ \end{align}$
por lo tanto,
$\begin{align}\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx&=\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx\\ &=\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx-\dfrac{\pi}{2}\ln 2-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx \end{align}$
Por lo tanto,
$\begin{align}\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx&=\dfrac{1}{2}\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx-\dfrac{\pi}{4}\ln 2\\ &=-\dfrac{G}{2}-\dfrac{\pi}{4}\ln 2\\ \end{align}$
Por lo tanto,
$\begin{align} \int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx=\dfrac{G}{2}+\dfrac{\pi}{8}\ln 2\\ \end{align}$
El resultado es el siguiente,
$\boxed{J=-\dfrac{G}{4}+\dfrac{1}{24}+\dfrac{\pi}{64}-\dfrac{\pi}{16}\ln2}$
Roger Hoover
Puntos
56
Es fácil comprobar que $\int_{0}^{1}x^2\sqrt{1-x^2}\,dx=\frac{\pi}{16}$ .
Al establecer $x=\cos\theta$ el problema se reduce a la computación
$$ \int_{0}^{\pi/2}\cos^2(\theta)\sin^2(\theta)\log\sin^2\frac{\theta}{2}\,d\theta$$ que es simple a través de la serie de Fourier de $\log\sin$ y $\log\cos$ .
