Lebesgue measure and limit of the integral.

Question

I have problem with this integral and limit : $\lim\limits_{n \to \infty } \int_{\mathbb{R}^+} \left( 1+ \frac{x}{n} \right) \sin ^n \left( x \right) d\mu_1$. Where $\mu_1$ is Lebesgue measure. I know, that I have to show that $(1+\frac{x}{n})\sin ^n(x)$ is bounded, yes?

Answer 1

When $n$ is odd, $\sin^n x$ is periodic of period $2\pi$, is $\ge 0$ on $[2k\pi,(2k+1)\pi]$ and is $\le 0$ on $[(2k+1)\pi, 2(k+1)\pi]$ for any integer $k$.

Letting $f(x) = \left(1+\frac xn\right)\sin^n x$, by definition the integral of $f$ is$$\int_{\Bbb R_+} f\,d\mu := \int_{\Bbb R_+} f_+\,d\mu - \int_{\Bbb R_+} f_-\,d\mu$$ where $f_+(x) := \max\{f(x), 0\}$ and $f_-(x) := \max\{-f(x), 0\}$

But $$\begin{align}\int_{\Bbb R_+} f_+\,d\mu &= \sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\left(1+\frac xn\right)\sin^n x\,dx\\ &\ge \sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\sin^n x\,dx\\ &= \sum_{k=0}^\infty\int_0^{\pi}\sin^n x\,dx\\&=\infty\end{align}$$ Similarly, $$\begin{align}\int_{\Bbb R_+} f_-\,d\mu &= \sum_{k=0}^\infty\int_{(2k+1)\pi}^{2(k+1)\pi}-\left(1+\frac xn\right)\sin^n x\,dx\\ &\ge \sum_{k=0}^\infty\int_{(2k+1)\pi}^{2(k+1)\pi}-\sin^n x\,dx\\ &= \sum_{k=0}^\infty\int_{\pi}^{2\pi}-\sin^n x\,dx\\&=\infty\end{align}$$

So $$\int_{\Bbb R_+} f\,d\mu = \infty - \infty$$

Which is undefined. Since you are trying to take the limit of an undefined sequence, the limit is also undefined.