Improper integral involving $\frac{\sin{nt}}{t}$ Assume $f(t) \colon [0,1] \to \mathbb{R}$ is a smooth function with $f(0) = 1$. Find the value of $\int_0^1\frac{\sin{nt}}{t}f(t)\,\mathrm{d}t$ as $n$ approaches infinity.
I've tried approaching this integral in several ways but have made no headway. 
I understand that this may have something to do with the fact that the integral of $\frac{\sin{t}}{t}\mathrm{d}t$ converges.
 A: We can write 
$$f(t)=1+\int_0^tf'(s)\,ds=1+t\int_0^1f'(yt)\,dy,$$
hence
$$\int_0^1\frac{\sin(nt)}tf(t)\,dt=\int_0^1\frac{\sin(nt)}t+\int_0^1\sin(nt)\int_0^1f'(yt)\,dy\,dt.$$
Since $f$ is twice continuously differentiable, the map $t\mapsto \int_0^1f'(yt)\,dy$ is of class $C^1$, and by Riemann-Lebesgue lemma or an integration by parts, we have
$$\lim_{n\to +\infty}\int_0^1\sin(nt)\int_0^1f'(yt)\,dy\,dt=0.$$
For the first term, do the substitution $s=nt$, $dt=\frac{ds}n$ to get 
$$\int_0^1\frac{\sin(nt)}t\,dt=\int_0^n\frac{\sin s}{\frac sn}\frac{ds}n=\int_0^n\frac{\sin s}s\,ds.$$
It converges to $\frac{\pi}2$, see here. 
A: Since $f$ is continuous, $f(x)\to f(0)$ as $x\to 0$.  This implies that for every $\varepsilon>0$, no matter how small, if $n$ is big enough, then $f\left(\dfrac u n\right)$ is between the points $f(0)\pm \varepsilon$.
$$
\begin{align}
& \int_0^1 \frac{\sin nt}{t} f(t) \, dt = \int_0^1 \frac{\sin nt}{nt} f\left(\frac{nt}{n}\right) \Big( n\,dt\Big) \\[10pt]
& = \int_{t=0}^{t=1} \frac{\sin u}{u} f\left(\frac{u}{n}\right) \, du = \int_{u=1}^{u=n} \frac{\sin u}{u} f\left(\frac{u}{n}\right) \, du\tag{1}
\end{align}
$$
If all of the values of $g(u)$ are between $f(0)\pm\varepsilon$ then
$$
\int_0^n \frac{\sin u}{u} g(u)\,du\text{ is between }\int_0^n f\left(\frac{\sin u}{u}\right) (f(0)\pm\varepsilon)\, du =(1\pm\varepsilon)\int_0^n\frac{\sin u}{u}\,du.\tag{2}
$$
We haven't yet established that the last expression in $(1)$ approaches a limit as $n\to\infty$, but it has a liminf and a limsup.  The last integral in $(2)$ does have a limit $L$ as $n\to\infty$.  What we've done entails that the liminf and limsup of the last expression in $(1)$ are between $(1\pm\varepsilon)L$.  That liminf and that limsup are quantities that don't depend on $n$.  And, for every $\varepsilon$, no matter how small, that liminf and that limsup must be between $(1\pm\varepsilon)L$.  That can happen only if the liminf and the limsup are both equal to $L$.
A: hint: $\frac {\sin(nt)}{\pi t}\to \delta(t)\ $ as $n\to\infty\ $ (see here)
(here we have only half of it!)
