Q:

How to calculate Fourier transform of this function?

the function is given by $$f(t)=\dfrac{1}{e^{at}-1}$$
Now, i know from the definition i've to evaluate this integral $$\begin{align}\mathcal{F}[f(t)]&=\int_{0}^{\infty}e^{-j \omega t}\left(\dfrac{1}{e^{at}-1}\right)\,dt \\&=\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{-t(j\omega+an)}\,dt \\&=\sum_{n=1}^{\infty}\dfrac{1}{j\omega+an}\end{align}$$
Is it the right way to approach ? 

A:

As far as I can tell, the integral diverges, but here's how one can attack the problem using known Fourier Transform pairs and Fourier Transform theorems.
I assume $a > 0$, $H(t)$ is the Heavyside unit step function, and I'll use the substitution $\omega = 2\pi s$ to use the Fourier Transform form with which I'm most comfortable.
$$I = \int_{0}^{\infty} e^{-2\pi ist}\left(\dfrac{1}{e^{at}-1}\right)dt$$
$$= \dfrac{1}{2} \int_{0}^{\infty} \dfrac{1}{e^{\frac{a}{2}t}}\cdot\dfrac{2}{e^{\frac{a}{2}t}-e^{-\frac{a}{2}t}}\cdot e^{-2\pi i st}dt$$
$$= \dfrac{1}{2} \int_{-\infty}^{\infty} H\left(\frac{a}{2}t\right) e^{-\frac{a}{2}t}\text{cosech}\left(\frac{a}{2}t\right)e^{-2\pi ist}dt$$
$$= \dfrac{1}{2} \mathscr{F}\left\{H\left(\frac{a}{2}t\right) e^{-\frac{a}{2}t}\right\}*\mathscr{F}\left\{\text{cosech}\left(\frac{a}{2}t\right)\right\}$$
Looking up the transform pairs in The Fourier Transform and Its Applications, by Ronald N. Bracewell, amd applying some Fourier Transform theorems found in that same text, we have
$$= \dfrac{1}{2}\dfrac{2}{a}\left[\dfrac{1-i2\pi s \frac{2}{a}}{1+\left(2\pi s\frac{2}{a}\right)^2}\right]*-i\dfrac{2\pi}{a}\tanh\left(\pi s \frac{2\pi}{a}\right)$$
Which after a bit of algebra and substituting $2\pi s = \omega$ yields
$$= -\left(\dfrac{\pi}{a}\right)^2\left(\dfrac{1}{\dfrac{\pi}{a}\omega -i\dfrac{\pi}{2}}\right) * \tanh\left(\dfrac{\pi}{a}\omega \right)$$
With a little thought, it's easy (for me at least) to see that the above convolution diverges. 

