Let $H$ be a complex hilbert space and $u,v \in H$. I need to find the integral:
$$\int_0^1 ||u+e^{i2\pi\theta}v||d\theta$$
I think, I should start like this: $$\int_0^1 ||u+e^{i2\pi\theta}v||d\theta = \int_0^1 \bigg( \Vert u \Vert^2+|-e^{i2 \pi \theta}| \Vert v \Vert^2 - 2Re \langle u,-e^{i2 \pi \theta} v \rangle \bigg) d\theta\\ = \Vert u \Vert^2 + \Vert v \Vert^2 -2 \int_0^1 Re \langle u,-e^{i2 \pi \theta} v \rangle d\theta$$
However, I am not quite sure how to continue from this point. Thought about using polarization identity: $$Re \langle x,y \rangle = \frac{1}{4} \big( \Vert x+y\Vert^2 - \Vert x-y\Vert^2 \big)$$
Any ideas?
Assuming your inner product is linear in its right-hand argument (sorry, I learned bra-ket notation in the context of quantum mechanics), it should be $$\int_0^1\left(\Vert u\Vert^2+\Vert v\Vert^2+e^{2\pi i\theta}\langle u,\,v\rangle+e^{-2\pi i\theta}\langle v,\,u\rangle\right)d\theta.$$If you use the opposite convention, the exponents proportional to $\theta$ will change sign. But they integrate out here anyway, so the final result is $\Vert u\Vert^2+\Vert v\Vert^2$.