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I have a problem with the following exercise. I don't really have an idea where to start. I'm glad about every help. So here is the exercise:

Suppose $f\colon R \to R$ is a function in $L^1(R)$ (i.e integrable function). Show that $\lim_{n \to \infty}\int f(x)\sin(nx)dx=0.$

Thanks already for any help!

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    obviously $\int_a^b\sin(nx)dx\to0$. now approximate $f$ by simple functions (see http://planetmath.org/encyclopedia/RiemannLebesgueLemma.html)2011-12-14

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I'll assume you want to take definite integrals over $\Bbb R$ (see Arturo's comment).

Given $\epsilon>0$, there is an $M$ such that $\int_{[-M,M]^c} |f(x)\sin(nx)|<\epsilon$ for all positive integers $n$. From this, it follows that you need only show that $\lim\limits_{n\rightarrow\infty}\int_{-M}^M f(x)\sin(nx) =0 $ for any fixed $M$.

To do this, you can appeal to the theorem cited by yoyo in the comments; or, you can use the hints in this very similar post

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    @Kurome You could apply the Dominated Convergence Theorem to the sequence $(g_n)$ with $g_n =f\cdot\chi_{[-n,n]}$.2016-05-17
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Here's an answer assuming that you meant to write $\sin \left( \frac{x}{n}\right)$:

You can use the Lebesgue dominated convergence theorem:

You have $|f(x) \sin(\frac{x}{n})| \leq |f(x)|$ which you know is integrable hence you can swap the limit and the integral:

$ \lim_{n \to \infty} \int f(x) \sin (\frac{x}{n})\ dx = \int f(x) \lim_{n \to \infty} \sin (\frac{x}{n})\ dx = \int f(x) \cdot 0\ dx = 0.$