Exists Lebesgue integral Lebesgue integral $\displaystyle\int^0_1 \log x \log(1+x) dx$ exists?
I know that
$2(x-1)\leq \log x\leq x-1$ for $1/2<x<1$
$x\leq |\log(x-1)|\leq 2x$, for $0<x<1/2$
Then $|\log\log(x+1)|\leq 2x|\log x|+4(1-x)|\log(1-x)|$
 A: The only problem with the integrand happens at near $0$. Since $\lim_{x\rightarrow0}\sqrt{x}\log(x)=0$ We get that
$$|\log(x)\log(1+x)|\leq \frac{M}{x^{1/2}}$$
for some $M>0$ and all $0<x\leq 1$.
From there you see that your integral indeed exists.
As it has been pointed out by @Quanto, you can then use standard tricks to evaluate the inetegral.
A: Yes, the lebesgue integral exists. To check it you just have to prove that the function is bounded in the interval, because a measurable, bounded function in a space of finite measure is always integrable.
That can be checked by calculating the limits of the function in $0$ and $1$ using L´Hôpital´s rule (both limits are 0).
A: It can actually be integrated explicitly as follows
\begin{align} \int^0_1 \log x \log(1+x) dx &\overset{IBP}=
\int_0^1 \left( \ln(1+x) + \ln x-\frac{\ln x}{1+x} \right)dx\\ &=2 \ln2 -2+ \frac{\pi^2}{12}
\end{align}
where the result $\int_0^1 \frac{\ln x}{1+x} = -\frac{\pi^2}{12}$ is used. Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$
