Tal vez sea demasiado pedir, pero hay una manera de solucionar $\int \limits_{0}^{1} \log(x)\log(1-x) dx$ sin convolución?
Tenga en cuenta que $\log x =\log_e x$.
Respuestas
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Paso 1: vamos a comprobar que para cada $n\in \mathbb{N}$, $$\int_{0}^{1} x^n \log(x) \, dx = -\frac{1}{(n+1)^2}$$ Starting with the identity $$\int_{0}^{1} x^n \, dx = \frac{1}{n+1}$$ differentiate both sides with respect to $$ n para obtener el resultado deseado.
Paso 2: Ampliación $\log(1-x)$ como su serie de Taylor, tenemos $$\log(x)\log(1-x) = -x\log(x) - \frac{x^2 \log(x)}{2} - \frac{x^3 \log(x)}{3} - \cdots$$ Sweeping issues of convergence under the rug, integrating both sides over the interval $[0,1]$ gives $$\int_{0}^{1} \log(x) \log(1-x) \, dx = \frac{1}{2^2} + \frac{1}{2\cdot 3^2} + \frac{1}{3\cdot 4^2} + \cdots$$ Paso 3: debemos evaluar la suma $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2} \right)$$ $$ = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) - \sum_{k=1}^{\infty} \frac{1}{(k+1)^2} = 1 - \left( \frac{\pi^2}{6} - 1\right) = 2-\frac{\pi^2}{6}$$
David H
Puntos
16423
La integración por partes nos encontramos con:
$$\begin{align} \mathcal{I} &=\int_{0}^{1}\ln{\left(x\right)}\ln{\left(1-x\right)}\,\mathrm{d}x\\ &=\left[x\ln{\left(x\right)}\ln{\left(1-x\right)}\right]_{0}^{1}-\int_{0}^{1}x\cdot\frac{d}{dx}\left[\ln{\left(x\right)}\ln{\left(1-x\right)}\right]\,\mathrm{d}x\\ &=-\int_{0}^{1}x\left[\frac{\ln{\left(1-x\right)}}{x}-\frac{\ln{\left(x\right)}}{1-x}\right]\,\mathrm{d}x\\ &=\int_{0}^{1}\left[\frac{x\ln{\left(x\right)}}{1-x}-\ln{\left(1-x\right)}\right]\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{x\ln{\left(x\right)}}{1-x}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1-x\right)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\ln{\left(x\right)}}{1-x}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1-x\right)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\ln{\left(x\right)}}{1-x}\,\mathrm{d}x-2\int_{0}^{1}\ln{\left(x\right)}\,\mathrm{d}x\\ &=2+\int_{0}^{1}\frac{\ln{\left(x\right)}}{1-x}\,\mathrm{d}x\\ &=2+\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{t}\,\mathrm{d}t;~~~\small{\left[1-x=t\right]}\\ &=2-\operatorname{Li}_{2}{\left(1\right)}\\ &=2-\frac{\pi^2}{6},\\ \end{align}$$
donde $\operatorname{Li}_{2}{\left(z\right)}$ es el dilogarithm función.
