Convolution of functions in the time domain is equivalent to multiplication in the frequency domain. However, I am interested in multiplication of functions in the time domain, which is convolution in the frequency domain.

For example, consider the constant function $f(t) = 1$. I would like to evaluate $f(t)^{2}$, but in the frequency domain. Using the convolution theorem, this is equivalent to $\frac{1}{2\pi i}F(s) * F(s)$. Since $F(s) = \frac{1}{s}$, this can be written as (disregarding the constant factor) $$ \large{\int_{-\infty}^{s} F(\zeta)F(s-\zeta)d\zeta} = \large{\int_{-\infty}^{s} \frac{d\zeta}{\zeta (s-\zeta)}}$$

Using partial fraction expansion, the rational function becomes $$ \large{ \frac{1}{s\zeta} + \frac{1}{s(s-\zeta)} } $$

So the integral reduces to $$ \large{ \frac{1}{s} \int_{-\infty}^{s} \frac{1}{\zeta}} + \frac{1}{s-\zeta} d\zeta = \frac{1}{s}(\log(\zeta) + \log(s-\zeta)) |_{-\infty}^{s}$$

I am not sure how to evaluate this integral. Are my limits incorrect since the variable is complex, or is there something else I missed?

  • $\begingroup$ Your convolution integral is wrong. Find the correct one in this table. $\endgroup$
    – Matt L.
    Jan 15, 2016 at 21:32
  • $\begingroup$ $$\frac{1}{2 i \pi} F \ast F(\sigma+it) = \int_{-\infty}^\infty F(\sigma+2i \zeta)F(\sigma+it-2i \zeta) d\zeta$$ see the convolution formula for Fourier transforms (the Laplace transform $F(\sigma+2i\pi\zeta)$ with $\zeta \in \mathbb{R}$ being the Fourier transform of $f(x) e^{-\sigma x}$) $\endgroup$
    – reuns
    Jan 15, 2016 at 21:43


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