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For $f \in \mathcal{S}(\mathbb R)$, where $\mathcal{S}$ is a Schwartz space, denote its Fourier transform as

$$ \hat{f}(\omega) := \mathcal{F}f(t) := \int_{-\infty}^{\infty} f(t) e^{-2 \pi i \omega t} dt $$ and corresponding inverse Fourier transform as $$ \mathcal{F}^{-1} \hat{f}(\omega) := \int_{-\infty}^{\infty} e^{2 \pi i t \omega} \hat{f}(\omega) d\omega $$

I would like to know what is the formula for:

$$ \mathcal{F} \left[ \int_{-\infty}^t f(\tau) d\tau \right] $$

What I have so far is:

\begin{alignat}{2} f(\tau) &= \mathcal{F}^{-1}\hat{f}(\omega) \qquad \implies \\ \int_{-\infty}^t f(\tau) d\tau &= \int_{-\infty}^t \mathcal{F}^{-1}\hat{f}(\omega) d\tau \\ &= \int_{-\infty}^t \left[ \int_{-\infty}^\infty e^{2 \pi i \tau \omega} \hat{f}(\omega) d\omega \right] d\tau \\ &= \int_{-\infty}^\infty \hat{f}(\omega) \int_{-\infty}^t e^{2 \pi i \tau \omega} d\tau\, d\omega \\ &= \int_{-\infty}^\infty \hat{f}(\omega) (2 \pi i \omega)^{-1}\left[ e^{2 \pi i \tau \omega} \right]_{-\infty}^t d\omega \\ \end{alignat}

and then I have a problem that the limit $$ \lim_{\tau \to -\infty}e^{2 \pi i \tau \omega} $$ is not well defined.

I wonder how did they get the corresponding result on this page:

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    $\begingroup$ That's not the correct definition for the Fourier transform of $L^2(\mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $\mathcal S(\mathbb R^n)$ and $L^1(\mathbb R^n)$ functions and then, since $\mathcal S(\mathbb R^n)$ is dense in $L^2(\mathbb R^n)$ you extend the Fourier transform operator by continuity. $\endgroup$
    – rubik
    Feb 22, 2017 at 9:42
  • $\begingroup$ ok, let me edit. $\endgroup$
    – aberdysh
    Feb 22, 2017 at 9:45

2 Answers 2

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Simple way is to calculate the Fourier transform of the distribution $\theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $\theta (x)$ by $\theta(x) e^{-\epsilon x}$ and take the limit $\epsilon \to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.

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  • When $f \in S(\mathbb{R})$ and $\int_{-\infty}^\infty f(t) dt = 0$ then $\int_{-\infty}^t f(\tau)d\tau \in S(\mathbb{R})$.

    Integrating by parts $$\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau](\xi) = \int_{-\infty}^\infty\int_{-\infty}^t f(\tau) d\tau e^{-2i \pi \xi t} dt = \int_{-\infty}^\infty f(t) \frac{e^{-2i \pi \xi t}}{2i \pi \xi} dt = \frac{1}{2i \pi \xi} \mathcal{F}[ f(t)](\xi)$$

  • Otherwise $\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau](\xi) = \frac{1}{2i \pi \xi} \mathcal{F}[ f(t)](\xi)+ \frac{\int_{-\infty}^\infty f(t) dt}{2}\delta(\xi)$$

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  • $\begingroup$ I don't see where $ \frac{\int_{-\infty}^\infty f(t) dt}{2}\delta(\xi)$ comes from $\endgroup$
    – aberdysh
    Feb 22, 2017 at 22:51
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    $\begingroup$ @aberdysh it comes from the Fourier transform of distributions. Let $H(\xi) = \mathcal{F}[\int_{-\infty}^t f(\tau)d\tau](\xi)$ then using the properties of the Fourier transform of distributions $2i \pi \xi H(\xi)= \mathcal{F}[f(t)](\xi)$ i.e. $H(\xi) = \frac{1}{2i \pi \xi} \mathcal{F}[f(t)](\xi)+C \delta(\xi)$ for some $C$ $\endgroup$
    – reuns
    Feb 22, 2017 at 23:54

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