For $f \in \mathcal{S}(\mathbb R)$, where $\mathcal{S}$ is a Schwartz space, denote it's 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 got the corresponding result on this page:

  • $\begingroup$ 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 '17 at 9:42
  • $\begingroup$ ok, let me edit. $\endgroup$ – aberdysh Feb 22 '17 at 9:45

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.

  • 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 $$\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau](\xi) = \frac{1}{2i \pi \xi} \mathcal{F}[ f(t)](\xi)+ \frac{\int_{-\infty}^\infty f(t) dt}{2}\delta(\xi)$$

  • $\begingroup$ I don't see where $ \frac{\int_{-\infty}^\infty f(t) dt}{2}\delta(\xi)$ comes from $\endgroup$ – aberdysh Feb 22 '17 at 22:51
  • $\begingroup$ @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 '17 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.