I have to find the convolution of the functions $f$ and $g$ on the interval $(-\pi, \pi)$. $$ f(x) = x+2;\ \ \ g(x) =\frac{\sin\frac{7x}{2}}{\sin\frac{x}{2}} $$
Here is my attempt: $$ \begin{aligned} &(f*g)(x)=\int\limits_{-\pi}^{\pi}f(t)g(x-t)dt=\int\limits_{-\pi}^{\pi}(2+t)\cdot\frac{\sin\frac{7(x-t)}{2}}{\sin\frac{x-t}{2}}dt=I_1+I_2\ \text{where}\\ &I_1=2\int\limits_{-\pi}^{\pi}\frac{\sin\frac{7(x-t)}{2}}{\sin\frac{x-t}{2}}dt=4\cdot\int\limits_{\frac{-\pi-x}{2}}^{\frac{\pi-x}{2}}\frac{\sin 7z}{\sin z}dz\\ &I_2=\int\limits_{-\pi}^{\pi}t\cdot\frac{\sin\frac{7(x-t)}{2}}{\sin\frac{x-t}{2}}dt \end{aligned} $$ The problem is to evaluate $I_1$ and especially $I_2$. Perhaps there is an easier way of cracking this problem?
You can first evaluate Fourier series for $f(x)$ and then use its relation to the Dirichlet kernel:$$S_n(x)=\int_{-\pi}^{\pi}f(x)D_n(x-t)dt$$ where $D_n(t)=\frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})}$ and $S_n(x)$ is n-th partial sum of Fourier series. In your case, $g(t)=\frac{\sin\frac{7t}{2}}{\sin\frac{t}{2}}$ equals $2\pi \cdot D_3(t)$.