Show that the complex Fourier Series for $$f(x)=x , \lvert x \rvert \le2$$defined over one period, is given by
$$f(x) = \frac{j2}{\pi}\sum_{r=-\infty}^\infty \frac{(-1)^r}{r} \exp\left(j\frac{r \pi x}{2}\right)$$
I know that it amonts in solving an integral, but it did not coincide with the series. Solving this integral.
$$ c_n=\frac{1}{4} \int_{0}^2 te^{-(jn\pi t)/2} dt $$ Integrating by parts yielded into
$$c_n= \frac{1}{(n\pi)^2}(e^{-jn\pi}-1 ) -\frac{1}{jn\pi}(e^{-jn\pi})$$
However, $$e^{-jn\pi}=\cos(n\pi)-j\sin(n\pi) $$ For the "sin" term it's always zero because its multiple of $\pi$. So it can be boiled down into $$e^{-jn\pi}=(-1)^n$$ Where $c_n$ may differ depeding on whether n is even or odd. Now the problem is that it does not coincide with the series.Can anyone tell where is the mistake? Remark: j is equivalent to the imgainary number i.
You should have:
$$c_n=\frac{1}{4} \int_{-2}^2 te^{-(jn\pi t)/2} dt$$
Note the $-2$ lower bound. That is probably the main source of your error. For example, that is how you ended up with $\frac{1}{(n\pi)^2}$.
When $n=0$, we easily see that $c_0=\int_{-2}^{2} t\,dt = 0$.
Then, when $n\neq 0$, we integrate by parts, letting $u=t,dv=e^{-jn\pi t/2}\,dt$ then $du=dt$ and $v=\frac{2}{-jn\pi} e^{-jn\pi t/2}=\frac{2j}{n\pi}e^{-jn\pi t/2},$ since $\frac{1}{-j}=j$.
But:
$$\int_{-2}^{2} v\,du = \frac{2j}{n\pi}\int_{-2}^2 e^{-j\pi n/2} = 0$$
when $n\neq 0$.
So we are left with:
$$c_n = \frac{1}{4}\left(u(2)v(2)-u(-2)v(-2)\right)$$
But $u(2)=2, u(-2)=-2$ and you get:
$$c_n = \frac{1}{2}(v(2)+v(-2)) = \frac{1}{2}\frac{2j}{n\pi}\left(e^{-jn\pi}+e^{jn\pi}\right)$$
But $e^{jn\pi}=e^{-jn\pi}=(-1)^n$. So:
$$c_n = \frac{2j}{n\pi}(-1)^n$$