Assume $f$ satisfies the assumptions of Dirchlet's Theorem - i.e. $f$ is a piecewise continuous complex function that has one-sided derivatives at each point in $[-\pi,\pi]$. Determine the following limit:
\begin{equation} \lim_{n \to \infty}\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\sin{\left(nt-\frac{t}{2}\right)}dt\tag{*} \end{equation}
What I have tried
I first rewrote the sinus expression and used one of the trigonometric addition formulas
\begin{align} \sin{\left(nt-\frac{t}{2}\right)} &=\sin{\left( \left[n+\frac{1}{2} \right]-1 \right)t} \\ &=\sin{(n+\frac{1}{2})t}\cos{(nt)}-\sin(nt)\cos(n+\frac{1}{2})t \end{align}
This meant I could repress the integral (due to linearity) as:
\begin{align} \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\sin{\left(nt-\frac{t}{2}\right)}dt &=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\left[\sin{(n+\frac{1}{2})t}\cos{(nt)}-\sin(nt)\cos(n+\frac{1}{2})t \right]dt \\ &=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\sin(n+\frac{1}{2})t\cos(nt)dt \\ &-\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\cos(n+\frac{1}{2})t\sin(nt)dt \quad \quad \quad \quad \quad \quad \quad \quad \quad(**)\end{align}
Now, focusing on the first term only, I sought to \begin{align}\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\sin(n+\frac{1}{2})t\cos(nt)\frac{2\sin{\frac{t}{2}}}{2\sin{\frac{t}{2}}}dt &=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{f(t)}{t}\sin(n+\frac{1}{2})t\cos(nt)\frac{2\sin{\frac{t}{2}}}{2\sin{\frac{t}{2}}}dt \\ &=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)\frac{\sin(n+\frac{1}{2})t}{2\sin{\frac{t}{2}}}\frac{2\sin{\frac{t}{2}}}{t}dt \end{align}
I now note: \begin{equation} D_{m}(t)=\frac{\sin(n+\frac{1}{2})t}{2\sin{\frac{t}{2}}} \end{equation}
and \begin{equation} \lim_{t \to 0} \frac{2 \sin{\frac{t}{2}}}{t}=1 \end{equation}
Here is where I get stuck. What I want to do is to "force" the expression to have the appearance necessary for me to use a theorem that connects it to the partial sum of the Fourier series $f$; namely: \begin{equation} S_{m}(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x+t)D_{m}(t)dt \end{equation}
I have thought of defining $h(t):=f(t)\cos(nt)\frac{2\sin{\frac{t}{2}}}{t}$, then let $x$ tend to zero and use the facts that $\cos(0)=\frac{2\sin{\frac{t}{2}}}{t}=1$, so $h(0)=f(0)$, which by Dirchlet's theorem is given by. \begin{equation} \frac{f(0-)+f(0+)}{0}\end{equation}
Does this sound reasonable to you? Thanks!