$\lim_{n \to \infty} \int_a^b g(x) \sin^n(x)dx$

Question

I want to find $\lim_{n \to \infty} \int_a^b g(x) \sin^n(x)dx$ where $a<b$ and $g(x)$ is continuous.

I've been able to show that if $a=0$ and $b=1$, then the limit is $0$. But I'm not sure if this holds in general.

Is there a general proof that the limit is $0$? If so, what does it look like?

UPDATE: I should have mentioned that I can't use measure theory.

Answer 1

This may be cheating, but it is essentially a much less elegant proof via the other answers. Fix $\epsilon>0$, and denote the points where $\vert \sin(x) \vert=1$ by $\{ y_l \}_{l=1}^k\subset [a,b]$. Then $g(x)\cdot \sin^n(x)\rightarrow 0$ converges uniformly to $0$ on $[a,b]\setminus \bigcup (y_l- \frac {1}{2} \epsilon, y_l+ \frac{1}{2} \epsilon)$, therefore if you denote $E:=[a,b]\setminus \bigcup_{l=1}^k (y_l- \frac {1}{2} \epsilon, y_l+ \frac{1}{2} \epsilon)$, you have:

$\vert\int_{a}^{b}g(x)\sin^n(x)dx\vert \leq \sum_{l=1}^k\int_{y_l-\epsilon}^{y_l+\epsilon}\vert g(x) \sin^n(x)\vert dx+\int_E \vert g(x)\sin^n(x)\vert dx \leq$

$\overset{\vert \sin^n(x)\vert \leq 1}{\leq} \sum_{l=1}^k\int_{y_l-\epsilon}^{y_l+\epsilon}\vert g(x)\vert dx+ \int_E \vert g(x)\sin^n(x)\vert dx \leq$

$\overset{\vert g(x)\vert \leq M}{\leq} M\cdot\sum_{l=1}^k\int_{y_l-\epsilon}^{y_l+\epsilon}1 dx+\leq M\cdot \epsilon k + \int_E \vert g(x)\sin^n(x)\vert dx \rightarrow Mk \epsilon+0$.

where $M$ is an upper bound on $G$ obtained from continuity on $[a,b]$. Since $M$ and $k$ are fixed, and $\epsilon$ is arbitrary, this must imply that:

$\underset{n\rightarrow \infty}{\lim} \vert\int_{a}^{b}g(x)\sin^n(x)dx\vert \leq 0$.

There are issues on what the intervals that should be excluded are since right now, we could be integrating outside $[a,b]$, but this is less important than the underlining argument.

Essentially, you take intervals of length $\epsilon$ around where $\sin(x)=1$, and use uniform convergence on the complement. The integrand on these intervals is bounded, and the length of the intervals can be a small you want.

Answer 2

$\sin ^{n} x \to 0$ except at a countable number of points, hence almost everywhere. Dominated Convergence Theorem completes the proof.

Answer 3

You can just use the fact that :

$$ \ g(x) \sin^n(x) \underset{n\rightarrow\infty }{\longrightarrow} 0 \quad a.e$$

And you justify the fact that you can switch limits and integrals with the dominated convergence theorem (for example), the main argument being :

$$\ |g(x) \sin^n(x)| \leqslant |g(x)| \ a.e $$

with $|g|$ integrable on $[a,b]$.

Answer 4

Riemann integral and limits only:

If $[a,b] \cap \{n\pi + \pi /2\}_{n \in \Bbb Z} = \varnothing$, then $\sup_{[a,b]} |\sin(x)|<1$, hence $$ \overline {\lim_n} \left|\int_a^b g(x)\sin^n (x) \,\mathrm dx\right| \leqslant \overline{\lim_n} \int_a^b |g|(x) |\sin(x)|\, \mathrm dx \leqslant \overline{\lim_n} (b-a) \sup_{[a,b]} |g|\cdot (\sup_{[a,b]} |\sin(x)|)^n \to 0. $$

If $[a,b]\cap \{n\pi +\pi/2\}_{n \in \Bbb Z} \neq \varnothing$, then we compute the limit for the case when such intersection contains only 1 point. WLOG, suppose $|\sin (c)| = 1$ where $c \in (a,b)$, then for $0 < \delta < \min(b-c, c-a)$, \begin{align*} &\phantom{==}\overline {\lim_n} \left|\int_a^b g(x) \sin^n(x)\, \mathrm dx \right|\\ &\leqslant \overline {\lim_n} \int_a^{c-\delta} + \int_{c-\delta}^{c+\delta} + \int_{c+\delta}^b |g(x)||\sin(x)|^n\, \mathrm dx\\ &\leqslant \overline {\lim_n} \int_{[a,b]\cap \{|x-c|\geqslant \delta\}} |g(x)||\sin(x)|^n \,\mathrm dx + \overline {\lim_n} \int_{|x-c|\leqslant \delta} |g(x)||\sin(x)|^n \, \mathrm dx\\ &\leqslant (b-a-2\delta) \overline{\lim_n}\sup_{[a,b]} |g| \cdot \sup_{[a,c-\delta]\cup [c+\delta, b]} |\sin(x)|^n + \overline{\lim_n} 2\delta\sup_{[a,b]} |g|\\ &= 0 + 2\delta \sup_{[a,b]} |g|\\ &\to 0 \quad [\delta \to 0^+]. \end{align*}

For more questionable points, do the similar thing to split the small intervals around them.