If I use the following set of Fourier transforms:

$$\mathcal F(w) = \int_{0}^T dtf(t)e^{iwt},$$

$$ f(t) = \frac{1}{2\pi} \int_{-w_N}^{w_N} dw\mathcal F(w)e^{-iwt},$$

where $f(t)$ is an arbitrary time domain signal with total duration $T$, $w= 2\pi f$ is angular frequency, and $w_N$ is the Nyquist angular frequency. The underlying assumption here is that the signal $f(t)$ is casual in the sense that it has a zero value for $t \leq 0$.

Now, what is the inverse Fourier Transform of $\sqrt {-iw} \mathcal F(w)$:

${}$ $$\mathcal F^{-1} (\sqrt {-iw} \mathcal F(w))=\quad?$$


Any help is appreciated.

  • $\begingroup$ What is the context of this question? Why are you interested in this? $\endgroup$
    – msm
    Dec 21, 2016 at 6:26

1 Answer 1


Hint. Refer Table of Fourier/Inverse Fourier transform. And make it a good habit to write the differential after the integrand.

  • $\begingroup$ Fixed the link. It would be better if you included the relevant portions of the table here. $\endgroup$
    – Batman
    Dec 21, 2016 at 5:27
  • $\begingroup$ Could you please tell me which one can help me? $\endgroup$
    – S. Amani
    Dec 21, 2016 at 11:59

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