What is the limit of $\int_{\mathbb{R}}f(x)|\sin( \lambda x)| dx$ for all integrable functions $f:\mathbb{R} \rightarrow \mathbb{C}$? 
What is the limit of $$\int_{\mathbb{R}}f(x)|\sin( \lambda x)| dx$$ for all integrable functions $f\colon\mathbb{R} \rightarrow \mathbb{C}$ when $|\lambda| \rightarrow \infty$?

I tried to use the Lemma of Riemann-Lebesgue. I want to prove that the integral of $|\sin(\lambda x)|$ is zero over one period when $|\lambda| \rightarrow \infty$.
I tried this: $$\int_{\mathbb{R}}\mathbb{1}_{[0,\pi]}\sin(\lambda x) dx= \frac{\cos(\lambda 0)-\cos(\lambda \pi)}{\lambda}\xrightarrow{|\lambda|\to\infty} 0$$
But then I doubted if this was indeed correct because $\lambda$ can be negative.
Can anybody help me to prove this?
 A: Using the fact that $f\in L^1$ can be approximated in $L^1$ by a step function (a function which can be expressed as $\sum_{j=1}^n c_j1_{I_j}$, where $\{I_j\}_{j=1}^n$ is a finite collection of disjoint open intervals and $c_j\in \mathbb{C}$) makes the problem much easier.
Fix $\epsilon>0$. Write $f=g+h$, where $g=\sum_{j=1}^n c_j1_{I_j}$ is a step function and $\|h\|_{L^1}<\epsilon$.
You can probably observe
$$\forall j,\lim_{\lambda\rightarrow\infty}\int_{I_j}|\sin(\lambda x)|dx=\frac{2}{\pi}m(I_j)$$
(For instance, you can show that by ignoring the rightmost portion of $I_j$, whose contribution to the integral shrinks as $\lambda\rightarrow\infty$, to make the integral periodic.)
Adding up these, you get
$$\lim_{\lambda\rightarrow\infty}\int_{\mathbb{R}}g|\sin(\lambda x)|dx=\frac{2}{\pi}\int_{\mathbb{R}}gdx$$
Now, since
$$\lim_{\epsilon\rightarrow0}\sup_{\lambda\neq0}\left|\int_{\mathbb{R}}gdx-\int_{\mathbb{R}}fdx\right|=\lim_{\epsilon\rightarrow0}\sup_{\lambda\neq0}\left|\int_{\mathbb{R}}h|\sin(\lambda x)|dx\right|=0$$
, taking $\epsilon\rightarrow0$, the answer is
$$\frac{2}{\pi}\int_{\mathbb{R}} fdx$$
