How can an odd function have a Fourier coefficient $b_n$ be $0$?

Question

I'm trying to calculate Fourier coefficient $b_n$ of this periodic function:

I choose this subintervals:

$$f(t)= \left\{ \begin{array}{lcc} -A & if & \frac{-T}{4} \leq t < 0\\ A & if & 0 \leq t < \frac{T}{4}\\ 0 & if & \frac{T}{4} \leq t < \frac{3T}{4} \end{array} \right.$$

For calculate $b_n$

$$b_n = \frac{2}{T}\left(\int_{\frac{-T}{4}}^0 -A * \sin(nwt) dt + \int_{0}^{\frac{T}{4}} A * \sin(nwt) dt\right)$$

I try to calculate it manually and always I get $0$, and I think thay this is imposible because is an odd function, also if I put this formula in wolframalpha and I get the same, how it's that posible?

Answer 1

I put incorrectly the variables in the wolfram input and the limits,

$w = \frac{2\pi}{t}$ should be $w = \frac{2\pi}{t0}$

this is the correct answer formula

Also, this is the Fourier result solution

Answer 2

No, $b_n$ isn't $0$. And Wolfram Alpha confirms that. This is the result:

$$ \frac{4A}{n\omega}\sin^2\left(\frac{nT\omega}{8}\right). $$

Analitacally, take the first integral:

$$ \int_{-T/4}^0 -A\sin(n\omega t)dt = -\int_0^{-T/4} -A\sin(n\omega t)dt = \int_0^{-T/4}A\sin(n \omega t)dt. $$

Now make the change $t\rightarrow -s$: $dt\rightarrow ds$ and the limits are $0$ and $T/4$. So

$$ \int_0^{T/4} A\sin(-n\omega s)(-ds) = \int_0^{T/4} (-1)^2 A\sin(n\omega s)ds = \int_0^{T/4} A\sin(n\omega s)ds. $$

Thus

$$ b_n = \frac{4}{T} \int_0^{T/4} A\sin(n\omega t)dt = \frac{4A}{n\omega}\sin^2\left(\frac{nT\omega}{8}\right). $$