Calculate $\lim\limits_{n\rightarrow\infty} \int_0^\pi \sin(x)^{1/n}|\log(x)| dx$

Question

So we have been ask to calculate the limit: $$\lim\limits_{n\rightarrow\infty} \int_0^\pi \sin(x)^{1/n}|\log(x)| dx$$
Since the function $\sin(x)^{1/n}|\log(x)|$ is dominated by $|\log(x)| $ $\forall x\in(0,\pi) \forall n\in\mathbb N,$
the dominated convergence theorem applies and so:

$$\lim\limits_{n\rightarrow\infty} \int_0^\pi \sin(x)^{1/n}|\log(x)|dx$$ $$=\int_0^\pi \lim\limits_{n\rightarrow\infty} \sin(x)^{1/n}|\log(x)|dx$$$$=\int_0^\pi|\log(x)|dx$$
Is this right?

Answer 1

As I said a completely wrong thing in the comment, here is an answer.

You can indeed apply DCT on that interval as the primitive of $\log(x)$ is $x(\log(x)-1)$. Now we do a case division to deal with the absolute value and see that the integral evaluates to \begin{align}\int_0^\pi|\log(x)|dx&=-\int_0^1\log(x)dx+\int_1^\pi\log(x)dx\\ &=-x(\log(x)-1)\big|_0^1+x(\log(x)-1)\big|_1^\pi\\ &=2+\pi(\log(\pi)-1))<\infty\end{align}

So DCT is applicable and your reasoning applies