Calculus $L=\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx$

Question

Calculus: $$L=\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx$$

My tried:

I think that

We have $$\left|\sqrt[n]{x}\sin x\right|\leq 1$$

Hence

$$\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx=\int_{0}^{\pi}\sin x\: dx=2$$

Answer 1

$$ \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx \leq \sqrt[n]{\pi} \int_{0}^{\pi}\sin x\: dx=2 \sqrt[n]{\pi} $$

$\forall \delta$, $0\lt \delta \lt \pi$

$$ \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\geq \int_{\delta}^{\pi}\sqrt[n]{x}\sin x\: dx\geq \sqrt[n]{\delta } \int_{\delta }^{\pi}\sin x\: dx=\sqrt[n]{\delta } (1+\cos \delta)$$

so

$$\sqrt[n]{\delta } (1+\cos \delta)\leq \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\leq 2 \sqrt[n]{\pi}$$

Hence,

$$ 1+\cos \delta\leq\varliminf_{n\to \infty}\int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\leq \varlimsup_{n\to \infty}\int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\leq 2$$

Let $\delta\to 0$, we get

$$\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx=2$$

Answer 2

The dominated convergence theorem says that since there is a dominating function $g(x)= \pi$, that is $|x^{1/n} \sin x| \leq g(x)$ on $[0, \pi]$ for $n\geq 1$, and $\int_0^\pi g(x) dx < \infty$, we can pass the limit through:

$$\lim_{n \to \infty} \int_0^\pi x^{1/n} \sin x dx = \int_0^\pi (\lim_{n \to \infty} x^{1/n} \sin x) dx = \int_0^\pi \sin x dx = 2.$$