91Ó°ÊÓ

The given function is complex exponentials $e^{inx}$.

In terms of an infinite sum of sines and cosines, a Fourier series is an expansion of a periodic function.

The orthogonality relationships of the sine and cosine functions are used in Fourier series.

The $c_0$ coefficient is shown below.

$$\begin{gathered} c_0=\frac{1}{2\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}xdx \\ {c}_0=0 \end{gathered}$$

The other coefficients are as follows:

$$\begin{gathered} c_n=\frac{1}{2\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}x{e}^{-inx}dx \\ {c}_n=\frac{1}{2\mathrm{π}}{\left[-\frac{x}{in}{e}^{-inx}+\frac{1}{n^2}{e}^{-inx}\right]}_{-\mathrm{π}}^{\mathrm{π}} \\ {c}_n=\frac{1}{2\mathrm{π}}\left(-\frac{\mathrm{π}}{in}{e}^{-in\mathrm{π}}-\frac{\mathrm{π}}{in}{e}^{in\mathrm{π}}+\frac{1}{n^2}\left(e^{-in\mathrm{π}}-e^{in\mathrm{π}}\right)\right) \\ {c}_n=\frac{1}{2\mathrm{π}}\left[-\frac{\mathrm{π}}{in}2\cos \left(n\mathrm{π}\right)-\frac{1}{n^2}2i\sin \left(n\mathrm{π}\right)\right] \end{gathered}$$

So, $c_n=\frac{i}{n}(-1{)}^n$.

Then the function is, $f\left(x\right)=1+i\left(\underset{n=-∞}{\overset{∞}{∑}}{(-1)}^n\frac{e^{inx}}{n}\right)n≠0$.

For check, change $n→-n$.

$$\begin{gathered} f\left(x\right)=1+i\underset{n=-∞}{\overset{-1}{∑}}\frac{e^{inx}}{n}(-1{)}^n+i\underset{n=1}{\overset{∞}{∑}}{(-1)}^n \\ f\left(x\right)=1+ii\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{n}\left(e^{inx}-e^{-inx}\right) \\ f\left(x\right)=1-2i\underset{n=1}{\overset{∞}{∑}}{(-1)}^n\frac{\sin \left(nx\right)}{n} \end{gathered}$$

Therefore, the expansion is $f\left(x\right)=1+i\left(\underset{n=-∞}{\overset{∞}{∑}}{(-1)}^n\frac{e^{inx}}{n}\right)n≠0$.