# What is Inverse Fourier Transform of $2𝛼^2/(𝛼^2−𝜔^2)$?

What is the inverse Fourier transform of $$F(w) = \frac{4}{4+(j2\pi f)^2}$$?
I have two suggested solutions:

1. Assume $$j2\pi f = \omega$$ and use the standard transform $$e^{-\alpha |\tau |} = \frac{2\alpha}{\alpha ^2 + \omega ^2}$$ to obtain $$\frac{1}{4}e^{-2|\tau|}$$

2. Calculate $$f(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\alpha}{\alpha^2-\omega^2}e^{i\omega\tau}d\omega$$

For (2), how can I solve the integral in order to get the inverse Fourier transform $$f(\tau)$$?

In general Fourier transform (or inverse) take products into convolutions. Let $$F(x)$$ and $$G(x)$$ have transforms $$f(t)$$ and $$g(t)$$. Then the transform of $$F(x)G(x)$$ is $$h(t)=\int f(s)g(t-s)ds$$.

For your problem $$f(t)=\frac{1}{16}\int e^{-a|s|-b|t-s|}ds$$ where $$a=2$$ and $$b=8$$.

(Note: I may have constant wrong, but you see the point.)


It looks like there is some problem with the answer in "1." First note that it is indeed true that the Fourier Transform $$\fG(\omega)$$ of the function $$\fg(\tau)\eqd e^{-\alpha\abs{\tau}}$$ is $$\fG(\omega)\eqd\frac{2\alpha}{\alpha^2+\omega^2}$$. Assuming the definition of the Fourier Transform used is $$\fG(\omega)\eqd \int_{-\infty}^{\infty}\fg(\tau)e^{-j\omega\tau}\dtau$$, we can check the relation as follows: \begin{align} \fG(\omega) &\eqd \int_{-\infty}^{\infty}\fg(\tau)e^{-j\omega\tau}\dtau && \text{by definition Fourier Transform} \\&\eqd \int_{-\infty}^{\infty} e^{-\alpha\abs{\tau}} e^{-j\omega\tau}\dtau && \text{by definition of \fg(\tau)} \\&= \int_{-\infty}^{0} e^{-\alpha(-\tau)} e^{-j\omega\tau}\dtau + \int_{0}^{\infty} e^{-\alpha(\tau)} e^{-j\omega\tau}\dtau \\&= \int_{-\infty}^{0} e^{\tau(\alpha-j\omega)}\dtau + \int_{0}^{\infty} e^{\tau(-\alpha-j\omega)}\dtau \\&= \brlr{\frac{e^{\tau(\alpha-j\omega)}}{\alpha-j\omega}}_{-\infty}^{0} + \brlr{\frac{e^{\tau(-\alpha-j\omega)}}{-\alpha-j\omega}}_{0}^{\infty} && \text{by Fundamental Theorem of Calculus} \\&= \brs{\frac{1}{\alpha-j\omega} - 0} + \brs{0- \frac{1}{-\alpha-j\omega}} \\&= \brs{\frac{1}{\alpha-j\omega}}\brs{\frac{\alpha+j\omega}{\alpha+j\omega}} + \brs{\frac{1}{\alpha+j\omega}}\brs{\frac{\alpha-j\omega}{\alpha-j\omega}} && \text{(to rationalize the denominator)} \\&= \frac{\alpha+j\omega}{\alpha^2+\omega^2} + \frac{\alpha-j\omega}{\alpha^2+\omega^2} \\&= \boxed{\frac{2\alpha}{\alpha^2+\omega^2}} \end{align}

And so the Inverse Fourier Transform (IFT) of $$\frac{2\alpha}{\alpha^2+\omega^2}$$ is $$e^{-\alpha\abs{\tau}}$$.

However, your question does not ask for the IFT of $$\frac{2\alpha}{\alpha^2+\omega^2}=\frac{2\alpha}{\alpha^2+(2\pi f)^2}$$, but rather for the IFT of $$\frac{2\alpha}{\alpha^2+(j2\pi f)^2} = \frac{2\alpha}{\alpha^2+(j\omega)^2} =\frac{2\alpha}{\alpha^2-\omega^2}$$

• can you show me the steps to get the IFT? Feb 7 at 17:09
• I have tried to find the IFT but so far haven't been successful. I will work more on it though. Maybe someone else already knows the answer and could post it here? Feb 8 at 0:09
• WolframAlpha has something interesting to say. In essence ... $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4}{4-\omega^2}e^{i\tau\omega}d\omega = -\frac{i}{2}e^{-2i\tau}(-1+e^{4i\tau})sgn(\tau)$$ Feb 8 at 1:00
• \begin{align} \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4}{4-\omega^2}e^{i\tau\omega}d\omega &= -\frac{i}{2}e^{-2i\tau}(-1+e^{4i\tau})sgn(\tau) \\&= -\frac{i}{2}(-e^{-2i\tau}+e^{2i\tau})sgn(\tau) \\&= -\frac{i}{2}[2i\sin(2\tau)]sgn(\tau) && \text{by Euler's Identity} \\&= \sin(2\tau)sgn(\tau) \end{align} Feb 8 at 1:19
• Conjecture: IFT of $\frac{2\alpha}{\alpha^2-\omega^2}$ is $\sin(\alpha\tau)\mathrm{sgn}(\tau)$ $\ldots$ but I'm not sure at this point $\ldots$ Feb 8 at 1:25


I'll be honest with you---it looks like there is no Inverse Fourier Transform for $$2\alpha^2/(\alpha^2-\omega^2)$$. I'd be happy to be proven wrong...but at least consider the following...

(1) Let $$s\eqd j\omega$$. Then the Fourier Transform $$\int_{-\infty}^{\infty} \fg(\tau)e^{-j\omega\tau}\dtau$$ of $$\fg(\tau)$$ becomes the (Two-Sided) Laplace Transform $$\int_{-\infty}^{\infty} \fg(\tau)e^{-s\tau}\dS$$ of $$\fg(\tau)$$.

(2) Note that what we want is a function $$\fg(\tau)$$ with Laplace Transform $$\frac{2\alpha}{\alpha^2+s^2}$$. Why? Because then when we substitute back in $$s=j\omega$$ we would get $$\brlr{\frac{2\alpha}{\alpha^2+s^2}}_{s=j\omega} = \frac{2\alpha}{\alpha^2-\omega^2}$$ and the world would be a happier place.

(3) If only there was such a function $$\fg(\tau)$$ ... hmmmmm ... But wait! There is a function like that! In fact, if you happened to guess that the function $$\fg(\tau)$$ is $$\fg(\tau)\eqd 2\sin(\omega\tau)\step(\tau)$$ where $$\step(\tau)=1$$ for $$\tau\geq1$$ and $$0$$ otherwise (the step function), then you would be right! And you can check that here (where the One-Sided Laplace is used and so $$\step(\tau)$$ is implied).

(4) So the Inverse Laplace Transform of $$\fG(s)\eqd\frac{2\alpha}{\alpha^2+s^2}$$ is $$\fg(\tau)\eqd 2\sin(\omega\tau)\step(\tau)$$. So we just set $$s=j\omega$$ and we're all done here, right? Well not quite. That Laplace Transform $$\fG(s)$$ has a Region of Convergence of $$\Re(s)>0$$, where $$\Re(s)$$ is the real component of the complex $$s$$. The problem here is that the Fourier Transform exists along the imaginary axis of $$s$$---that is, at $$s=j\omega$$, which means $$\Re(s)=0$$ ... which is not in the Region of Convergence of $$\fG(s)$$ ... and so rather, it diverges or "goes to infinity" or "blows up" at $$s=j\omega$$.

And so, while the Laplace Transform of $$\fg(\tau)$$ does exist (for $$\Re(s)>0$$), it does not exist (it diverges) for $$\Re(s)\leq0$$. And so by extension, it looks like the Fourier Transform $$\fG(\omega)=\frac{2\alpha}{\alpha^2-\omega^2}$$ does not exist (because it is not in the Region of Convergence of $$\fG(s)$$).