Find $\inf _{a,b,c\in \mathbb C}\int _{[-\pi,\pi]}|x+a+b\cos x+c\sin x | ^2dx$

Question

I need to find $\inf _{a,b,c\in \mathbb C}\int _{[-\pi,\pi]}|x+a+b\cos x+c\sin x | ^2dx$. I thought about using the fact $\int _{[-\pi,\pi]}f\bar g$ is an inner product $ \left\langle f,g \right\rangle $ for which $1,\cos x,\sin x$ are orthogonal. Then I need to calculate $\inf_{a,b,c\in \mathbb C}\|x+a+b\cos x+c\sin x \|^2$. Using the orthogonality relations I have cut this down to $$\|x\|^2+2\operatorname{Re}(c) \overbrace{\left\langle x,\sin x \right\rangle}^{2\pi} +|a|^2\|1\|^2 +|b|^2 \|\cos x\|^2+|c|^2\|\sin x\|^2$$

So it seems like $a=b=c=0$ yields the infimum. On the other hand, this seems too simple to be true and I'm afraid I'm missing something.

What am I doing wrong?

Answer 1

Noting the oddness/evenness of the functions involved can lead us to:

$$\inf _{a,b,c\in \mathbb C}\int _{[-\pi,\pi]}|x+a+b\cos x+c\sin x | ^2dx=\inf _{c\in \mathbb C}\int _{[-\pi,\pi]}|x+c\sin x | ^2dx$$

Then:

$$\int _{[-\pi,\pi]}|x+c\sin x | ^2dx=\frac{2\pi^3}{3}+4\pi\Re{c}+\pi \vert c\vert ^2=\pi\vert c+2\vert ^2+\left(\frac{2\pi^3}{3}-4\pi\right)$$

which means that we minimise the integral at $a=b=0,c=-2$.