**Part 5(e):**We use an even continuation for this function. Integration was done using the angle-sum identity as in

Problem 3.

\begin{equation*}

b_n = 0.

\end{equation*}

\begin{equation*}

a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin{((m-\frac{1}{2})x)}dx \\

= \frac{-2}{\pi (m - \frac{1}{2}} \cos{(m-\frac{1}{2})x} \big|_{0}^{\pi}\\

=\frac{-2}{\pi (m- \frac{1}{2})}.

\end{equation*}

\begin{equation*}

a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin{((m-\frac{1}{2})x)}\cos{nx} dx\\

= -\frac{1}{\pi}\left[\frac{\cos{m + n - \frac{1}{2}} x }{m + n - \frac{1}{2}} + \frac{\cos{m - n - \frac{1}{2}} x }{m - n - \frac{1}{2}} \right]_{0}^{\pi}\\

= \frac{1}{\pi}\left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right).

\end{equation*}

\begin{equation*}

\sin{((m-\frac{1}{2})x)} = \frac{1}{\pi (m- \frac{1}{2})} + \frac{1}{\pi}\sum_{n=1}^{\infty} \left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right) \cos{n x}.

\end{equation*}