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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 ?

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  • $\begingroup$ You would have to argue why you can exchange sum and integral. Maybe look for something like dominated convergence or monotone convergence. $\endgroup$ – Jakob Elias May 22 '17 at 12:38
  • $\begingroup$ Are you familiar with contour integration and the residue theorem? Look at the likes of f(t)... $\endgroup$ – lux May 22 '17 at 13:29
  • $\begingroup$ yes i'm familiar with those , how to choose the contour in this case ? $\endgroup$ – Siddhartha Ganguly May 22 '17 at 13:36
  • $\begingroup$ I don't think this function is $L^1$ so a transform wouldn't be meaningful, even with the rewrite. $\endgroup$ – Sean Roberson May 22 '17 at 15:46
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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.

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  • $\begingroup$ Thanks @Andy Walls , it's a good approach, but wolfram gives this- wolframalpha.com/input/?i=fourier+transform+1%2F(e%5E(at)+-1) nice result $\endgroup$ – Siddhartha Ganguly May 24 '17 at 9:08
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    $\begingroup$ @Lelouch.D.Light Interesting. I'll have to go back and see if I can get that result as well. One subtle point is that your problem started with a one-sided Fourier transform, which is why I had the unit step function in the problem. It is unclear to me at the moment, if Wolfram Alpha is performing a 2 sided or 1 sided transform. I guess I'll find out. $\endgroup$ – Andy Walls May 24 '17 at 10:23

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