Fourier series and coefficients c1

Question

Let $f\in C_{st}$ be a function defined as: $$f(x) = e^{i\left|x\right|} \ \text{for $x\in]-\pi,\pi[$}$$

For $n\in\mathbb{Z}$ let $c_n$ define the $n^{th}$ Fourier Coefficient for $f$ with respect to the usual orthonormal system $\text{{${e_n |n\in\mathbb{Z}}$}}$

Show that $c_0 = \frac{2i}{\pi}$ og $c_1 = \frac{1}{2}$

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I have worked with complex Fourier Series where the $c_n$ coefficient is given by: $$c_n = \frac{1}{2\pi}\int_{\pi}^{\pi}f(x)e^{-inx}\mathrm{d}x.$$

Based on this, I believe I have found $c_0$

$$c_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i|x|}e^{-i0x}\mathrm{d}x=\frac{1}{2\pi}\int_{-\pi}^{0}e^{i(-x)}\mathrm{d}x+\int_{0}^{\pi}e^{ix}\mathrm{d}x=\frac{1}{2\pi}\left(i\left[e^{i(-x)} \right]_{-\pi}^0 +i \left[e^{ix} \right]_{0}^{\pi}\right)=\frac{1}{2\pi}\left(i\cdot(1+1 \right)+i(1+1)) =\frac{4i}{2\pi}=\frac{2i}{\pi}$$

However, I am not sure how to calculate $c_1$, I hope someone can show it.

Answer 1

Just work it out, like $c_0$.

$$c_1=\frac{1}{2\pi}\int_{-\pi}^{0}e^{i(-x)}e^{-ix}dx+\int_{0}^{\pi}e^{ix}e^{-ix}dx=\frac{1}{2\pi}\left[\left(\frac{e^{-i2x}}{-2i}\right)^{0}_{\pi}+\left(\pi\right)\right]=\frac{1}{2} $$ as desired.