Prove that Lebesgue Integral $\int_{0}^{1}\log\left(x\right)\log\left(x+1\right))dx$ exists.
I analyze if Riemann Instegral exists
$\int_{0}^{1}\log\left(x\right)\log\left(x+1\right))dx=[\log(x)((x+1)(\log(x+1)-1))]_{0}^{1}-\int_{0}^{1}\frac{(x+1)(\log(x+1)-1)}{x}dx$
using integration by parts but I don't know the value of $\int_{0}^{1}\frac{\log(x+1)}{x}dx$ and the first part of the integration by parts. I can't evaluate for $x=0$, could you tell me if this is the correct method to solve it?
Show that $\lim_{x \downarrow 0} (\log (1+x) \log(x)) = 0$. The function $f(x) = \log (1+x) \log(x)$ for $x>0$ and $f(x) = 0$ is continuous on $[0,1]$ hence measurable and bounded hence integrable.