$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$

Question

Demuestra: $ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$

Mi solución: $( x+1>x)$ $$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx< \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{1-x}}=0.5 $$

¿Existe una fórmula cerrada para las integrales?

Answer 1

Multiplica numerador y denominador por $$\sqrt[3]{(1+x)}^2 -\sqrt[3]{1+x}\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}$$

Answer 2

Una forma cerrada menos intimidante es: $$\frac5{12}\rho^2-\frac{\pi}{36\sqrt 3}-\frac{\sqrt 3}{18}\tan^{-1}\left(\frac{\rho^2+1}{3}\right)-\frac1{36}\ln\left(1+\frac{3\rho}{(\rho-1)^2}\right)$$ donde $\rho=2^{1/3}$.

Agregaré la demostración pronto.

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Prueba:

Multiplicando el numerador y el denominador por $(1+x)^{2/3}-(1-x^2)^{1/2}+(1-x)^{2/3}$,

$$\begin{align} I&=\frac12\int^1_0 x^{1/3}\left[(1+x)^{2/3}-(1-x^2)^{1/3}+(1-x)^{2/3}\right]dx \\ 2I&=\underbrace{\int^1_0 x^{1/3}(1+x)^{2/3} dx}_{A}-\underbrace{\int^1_0 x^{1/3}(1-x^2)^{1/3} dx}_{B}+\underbrace{\int^1_0 x^{1/3}(1-x)^{2/3} dx}_{C} \\ 2I&=A-B+C \end{align} $$

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Sea $\rho=2^{1/3}$. $$\begin{align} A&=\int^1_0 x^{1/3}(1+x)^{2/3} dx \\ &=\int^{1/2}_0\frac{w^{1/3}}{(1-w)^3}dw \qquad x=\frac{w}{1-w}\\ &=3\int^\rho_0\frac{y^3}{(1-y^3)^3}dy \end{align} $$

La calculadora integral dice $$A=\frac56\rho^2+\frac{\pi}{18\sqrt 3}-\frac{\sqrt 3}{9}\tan^{-1}\left(\frac{\rho^2+1}{3}\right)-\frac1{18}\ln\left(1+\frac{3\rho}{(\rho-1)^2}\right)$$

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$$\begin{align} B&=\int^1_0 x^{1/3}(1-x^2)^{1/3} dx \\ &=\frac12\int^1_0 y^{-1/3}(1-y)^{1/3}dy \qquad y=x^2 \\ &=\frac12\mathcal B\left(\frac23,\frac43\right) \\ &=\frac12\frac{\Gamma(\frac23)\Gamma(\frac43)}{\Gamma(2)} \\ &=\frac12\frac13\Gamma\left(\frac23\right)\Gamma\left(\frac13\right) \\ &=\frac16\frac{\pi}{\sin(\frac{\pi}3)} \\ &=\frac{\pi}{3\sqrt3} \end{align} $$

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De manera similar,

$$\begin{align} C&=\int^1_0 x^{1/3}(1-x)^{2/3} dx \\ &=\frac12\mathcal B\left(\frac43,\frac53\right) \\ &=\frac12\frac{\Gamma(\frac43)\Gamma(\frac53)}{\Gamma(3)} \\ &=\frac12\frac13\frac23\Gamma\left(\frac23\right)\Gamma\left(\frac13\right) \\ &=\frac19\frac{\pi}{\sin(\frac{\pi}3)} \\ &=\frac{2\pi}{9\sqrt3} \end{align} $$

Por lo tanto, $$I=\frac5{12}\rho^2-\frac{\pi}{36\sqrt 3}-\frac{\sqrt 3}{18}\tan^{-1}\left(\frac{\rho^2+1}{3}\right)-\frac1{36}\ln\left(1+\frac{3\rho}{(\rho-1)^2}\right)$$ lo cual es aproximadamente $0.43027$.