Let $a\in\mathbb{R}-\mathbb{Z}$. Why is the following equality true?
$$1 = \frac{1}{2\pi} \int_0^{2\pi} \left| e^{-i(\pi-x)a} \right|^2 dx$$
More precisely, why is the integrand equals $1$?
2026-05-15 09:21:39.1778836899
Why does the integral equal $1$?
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This is true for $a\in \mathbb Z$ too.
$-(\pi-x)a$ is real, so $-i(\pi-x)a$ is purely imaginary. It is well known that $$ e^{it}=\cos t + i\sin t \qquad\qquad\text{when }t\in\mathbb R$$ so its modulus is $\sqrt{\sin^2 t+\cos^2 t} = \sqrt 1 = 1$.