Fourier analysis, compute integral $\dfrac{1}{2\pi}\int_0^{2\pi}f(t)g(t)dt$ where $g(t)= \sum_{n=1}^\infty \dfrac{(-1)^n}{n}\sin(nt)$ and $f(t)=\cos(t)+7\sin(2t)$.
I'm having slight problems solving this integral, can we use alternating series test to say that the sum converges in some way? It feels like the sum might converge to $0$ and that we in that case will get an integal that is just $0$, but that feels counterintuitive so I'm unsure. Is there some theorem that might help that I'm missing?
The sum for $g$ converges in $L^2$ because $\sum_{n=1}^{\infty}\frac{1}{n^2} < \infty$ and because $\{ s_n(x)=\frac{\sin(nt)}{\sqrt{\pi}} \}_{n=1}^{\infty}$ is an orthonormal subset of $L^2[0,2\pi]$. That is, $$ g_N(t) = \sum_{n=1}^{N}\frac{(-1)^n}{n}\sin(nt) $$ converges in $L^2$ to some $g\in L^2$. Because of this, $$ \int_{0}^{2\pi}g(t)f(t)dt=\lim_{N}\int_{0}^{2\pi}g_N(t)f(t)dt \\ = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}\int_{0}^{2\pi}\sin(nt)f(t)dt \\ = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}\int_{0}^{2\pi}\sin(nt)(\cos(t)+7\sin(2t))dt. $$
The functions $\{ 1,\cos(t),\sin(t),\cos(2t),\sin(2t),\cdots \}$ are mutually orthogonal. So the above reduces to
$$ \frac{7}{2}\int_0^{2\pi}\sin^2(2t)dt=\frac{7\pi}{2}. $$ Therefore, $$ \frac{1}{2\pi}\int_{0}^{2\pi}f(t)g(t)dt = \frac{7}{4}. $$