I am wondering if there exists a closed form formula for the Fourier transform of the sigmoid function $f(x) = \frac{e^{x}}{(1 + e^{x}}$. More specifically I would need to calculate $F(w) = \int_{-\infty}^{\infty}(f(x) e^{-iwx} dx)$. This can be expressed as $F(w) = \int_{-\infty}^{\infty}(\frac{e^{x(1 – iw)}}{(1 + e^{x})}dx$. How would I go from here, or would there be a better way of finding the Fourier transform of the Sigmoid function?
I have tried obtaining a result using the in-built $fourier$ matlab function, it did however not find a solution.
Best Answer
Split the integral into two pieces:
$$\int_0^{+\infty} + \int_{-\infty}^0 \frac{e^{x(1 - i\omega)}}{1 + e^x}\ dx$$
Take the first integral, and collect $e^x$ at the denominator
$$\int_0^{+\infty} \frac{e^{x(1 - i\omega)}}{e^x(1 + e^{-x})} dx = \int_0^{+\infty} \frac{e^{-i\omega x}}{1 + e^{-x}}$$
Now you can use Geometric Series for the term
$$\frac{1}{1 + e^{-x}} = \sum_{k = 0}^{+\infty} (-1)^ke^{-kx}$$
Hence
$$\sum_{k = 0}^{+\infty} (-1)^k \int_0^{+\infty}e^{-x(k + i\omega)} dx = \sum_{k = 0}^{+\infty} (-1)^k \frac{1}{k + i\omega}$$
If you have a basic knowledge of special functions, this sum is called the generalized Lerch Phi function:
$$\Phi (-1,1,i \omega)$$
You can find tabulates everywhere.
For the second integral, due to the range, you don't collect anything, just use Geometric Series for
$$\frac{1}{1 + e^x} = \sum_{k = 0}^{+\infty} (-1)^k e^{kx}$$
hence
$$\sum_{k = 0}^{+\infty}(-1)^k \int_{-\infty}^0 e^{x(1 - i\omega + k)} = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{1 - i\omega + k}$$
Again a Lerch Phi function:
$$\Phi (-1,1,1-i w)$$
Eventually:
$$\Phi (-1,1,i \omega) + \Phi (-1,1,1-i \omega)$$
By mathematical manipulations, using the definition and special cases, that sum is nothing but
$$\frac{1}{2} i \pi \tanh \left(\frac{\pi \omega}{2}\right)-\frac{1}{2} i \pi \coth \left(\frac{\pi \omega}{2}\right)$$
Or more easily
$$-i \pi\ \text{csch}(\pi \omega)$$