This 3blue1brown video on Fourier series leaves the viewer with the following exercise:
$$c_n=\int_0^{0.5}e^{-2\pi int}\,\,\mathrm{d}t\,\,+\int_{0.5}^1-e^{-2\pi int}\,\,\mathrm{d}t\tag{1}$$
Show that
$$c_n=\frac{2}{n\pi i}\tag{2}$$
for odd $n$ and $c_n=0$ otherwise.
So what I first did is find the antiderivative of $e^{-2\pi int}$ which is
$$\frac{1}{-2\pi in}e^{-2\pi int}$$
Then the sum of the integrals $(1)$ can be written as
$$c_n=\left(\frac{1}{-2\pi in}e^{-\pi in}-\frac{1}{-2\pi in}e^0\right)+\left(-1\cdot\frac{1}{-2\pi in}e^{-2\pi in}+\frac{1}{-2\pi in}e^{-\pi i n}\right)$$
which can be simplified to
$$c_n=\frac{2e^{-\pi in}-1-e^{-2\pi i n}}{-2\pi i n}$$
From there on, I am clueless on how to show that this expression reduces to $(2)$ when $n$ is an odd number or to $0$ if it isn't. So what am I missing here, is my overall approach wrong?
I apologize if this question is trivial or stupid, but as a highschool student, I only have limited knowledge of things such as a Fourier series.
not bad for a high school student! Here is Euler's formula: $$e^{i \theta} = \cos \theta + i \sin \theta$$ Apply this to each of the exponential terms in your numerator. Then think about what are the values of $\sin \pi, \sin 2\pi, \ldots$ and $\cos \pi, \cos 2\pi, \ldots$