Complex Fourier series of $\sin(x)$, [$0,\pi$)

Question

I am trying to solve the following question:

I have tried to make an attempt at this question but I am not sure where to begin. My current understanding is to apply $f(x) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \sin(x)e^{-inx} dx$ to find the $c_n$ coefficient for the general formula of a complex Fourier series $\sum_{n = -\infty}^{\infty} c_ne^{inx}$.

Is this the correct approach to this question?

Any help, clarification or suggestions will be greatly appreciated so I begin to solve this question. Thank you!

EDIT: attempted to evaluate $\int sin(x)e^{-ix} dx$, as a start (dont worry i did not forget about $e^{-inx}$ to obtain $\frac{-ix}{2} - \frac{1}{4} e^{-2ix} + c$ but still not sure if this is correct.

Answer 1

$$\hat{f}(n) = \frac{1}{2\pi}\int_{0}^{2\pi} f(x) e^{-inx} dx$$

$$= \frac{1}{2\pi} \int_{0}^{\pi} \sin(x) e^{-inx} dx$$

Now writing $\sin x = \frac{e^{ix} -e^{-ix}}{2i}$, we can obtain $\hat{f}(n)$.

Once we have that we can write the Fourier series as:

$$\sum_{n=-\infty}^{\infty} \hat{f}(n) e^{inx}$$

Answer 2

An alternate solutions is using fourier relations as f.g -> F * G muliplication goes to convolution. Your function is sin multiplied by a top hat function convolved by a comb. The transform will thus be the two deltas from the sin function convolved by a sinc times by a comb (with a phase shift to compensate for the offset of the comb).