1. Integrating the Fourier coefficient

Consider a Fourier series of
$$-100S(x) = \sum_{n=1}^\infty b_n \sin(n\pi x)$$
where $S(x)$ is sigmoid function. The Fouerier coefficient is given by
$$b_n=2 \int_0^1 S(x) \sin(n\pi x)dx$$
Right?

If yes, what is the solution of this integral to calculate the Fourier coefficient for the above series?

2. Re: Integrating the Fourier coefficient

Yes, the sigmoid function is an odd function so has a Fourier sine series. The sigmoid function is $\frac{e^x}{e^x+ 1}$. The coefficient is given by $\int_0^1 \frac{e^x}{e^x+ 1}sin(n\pi x)dx$. I would integrate that using $sin(n\pi x)= \frac{e^{n\pi ix}- e^{-n\pi ix}}{2i}$ and combining the exponentials.

3. Re: Integrating the Fourier coefficient

Originally Posted by HallsofIvy
Yes, the sigmoid function is an odd function so has a Fourier sine series. The sigmoid function is $\frac{e^x}{e^x+ 1}$. The coefficient is given by $\int_0^1 \frac{e^x}{e^x+ 1}sin(n\pi x)dx$. I would integrate that using $sin(n\pi x)= \frac{e^{n\pi ix}- e^{-n\pi ix}}{2i}$ and combining the exponentials.
Thanks for the clarification, but I lost. You mean
$\int_0^1 \frac{e^x}{e^x+ 1}\frac{e^{n\pi ix}- e^{-n\pi ix}}{2i}dx$

Could you please elaborate how to integrate it?