# Removing $e^{-in\pi x/\ell}$ from an integral

I’m considering a proof of the convergence of the Fourier series. It begins by considering the full Fourier series of the periodic extension of $\phi$ defined on $[-\ell, \ell]$.

The full Fourier series is

$$\sum_{n=-\infty}^\infty C_ne^{\frac{in\pi x}\ell}, \quad C_n = \frac{1}{2\ell}\int_{-\ell}^\ell \phi(x)e^{\frac{-in\pi x}{\ell}}dx.$$

Then we attempt to bound $|C_n|$ as

$$|C_n| \leq \frac{1}{2\ell}\int_{-\ell}^{\ell} \left|\phi(x)e^{\frac{-in\pi x}{\ell}}\right| dx = \frac{1}{2\ell} \int_{-\ell}^\ell \left|\phi(x)\right|dx \leq \sup_{x \in [-\ell, \ell]} \left|\phi(x)\right|$$

I do not understand the justification for the first equality. Why can we drop the $|e^{\frac{in \pi x}{\ell}}|$ term from the integrand?

#### Solutions Collecting From Web of "Removing $e^{-in\pi x/\ell}$ from an integral"

For a real number $y$, $|e^{iy}| = 1$ since

$$|e^{iy}| = \sqrt{e^{iy}\overline{e^{iy}}} = \sqrt{e^{iy}e^{-iy}} = \sqrt{e^{i(y-y)}} = \sqrt{e^0} = \sqrt{1} = 1.$$

$|e^{it}| = 1$ if $t$ is real.