# Fourier transform of $\int_{-\infty}^{t} f(\eta )\text{d}\eta$

Suppose $f(t)$ and $F(\omega)$ are a Fourier transform pair. I want to show that $$\mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\} = \int_{-\infty}^t f(\eta)\ \text{d}\eta$$ I start with the Fourier transform of the RHS and use integration by parts: \begin{align*} \mathcal{F}\left\{\int_{-\infty}^tf(\eta)\ \text{d}\eta \right\}\ &= \int_{-\infty}^{\infty} \int_{-\infty}^t f(\eta)\ \text{d}\eta\ e^{-i\omega t}\ \text{d}t \\&=\underbrace{\frac{-1}{i\omega}\left[\int_{-\infty}^t f(\eta)\ \text{d}\eta\ e^{-i\omega t}\right]_{t\ =-\infty}^{t\ =\ \infty}}_{=\ 0 ?}\ +\ \frac{1}{i\omega}\int_{-\infty}^{\infty}f(t)\ e^{-i\omega t}\ \text{d}t\\ &= \frac{F(\omega)}{i\omega} \end{align*} Hence $$\int_{-\infty}^t f(\eta)\ \text{d}\eta\ = \mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\}$$

If the result is true I can see the first term on the RHS after performing the integration by parts must vanish but I'm not sure how to justify it. Any help on justifying it (or a cleaner approach to show the result) would be appreciated, thanks!

One may recall that

$$\mathcal{F}(g')(\omega)=i\omega\mathcal{F}(g)(\omega) \tag1$$

applying it with $$g(t)=\int_{-\infty}^t f(\eta)\ \text{d}\eta, \quad g'(t)=f(t),\quad$$ the notation $F(\omega):=\mathcal{F}(f)(\omega)$, it gives $$\frac{F(\omega)}{i\omega} =\mathcal{F}\left( \int_{-\infty}^t f(\eta)\ \text{d}\eta\right)(\omega) \tag2$$ that is

$$\mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\} = \int_{-\infty}^t f(\eta)\ \text{d}\eta. \tag3$$

Are you Ok with a proof of $(1)$ ?

• Très bon. :D No that's fine, I've already proved (1), thank you. – AtticusFinch95 Apr 27 '16 at 23:29
• You are welcome. – Olivier Oloa Apr 27 '16 at 23:30

Let $F(\omega)$ be the Fourier transform of the square integrable, continous function $f(t)$ as given by

$$F(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt$$

Let $I_L(\omega,)$ be defined by the integral

$$I_L(\omega)=\int_{-L}^L \int_{-\infty}^tf(t')\,dt'\,e^{-i\omega t}\,dt$$

Integrating by parts with $u=\int_{-\infty}^tf(t')\,dt'$ and $v=\frac{e^{-i\omega t}}{-i\omega}$ yields

$$I_L(\omega)=\frac{1}{i\omega}\int_{-L}^L f(t)e^{-i\omega t}\,dt+\frac{1}{i\omega}\left(e^{i\omega L}\int_{-\infty}^{-L}f(t')\,dt'-e^{-i\omega L}\int_{-\infty}^{L}f(t')\,dt'\right) \tag 1$$

Assuming that $\int_{-\infty}^t f(t')\,dt'$ is also a square integrable function, then we must have $$\lim_{L\to \infty}\int_{-\infty}^L f(t')\,dt'=0$$

Then, the boundary term on the right-hand side of $(1)$ vanishes in the limit as $L\to \infty$ and the limit of $I_L(\omega)$ becomes

\begin{align} \bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \left(\int_{-\infty}^t f(t')\,dt'\right)\,e^{-i\omega t}\,dt=\frac{1}{i\omega}F(\omega)}\end{align}

as was to be shown!

• Hello, thanks for your answer! Just a small typo on the last line, it should be $e^{-i\omega t}$. ;) Sorry, I'm kinda new to this so bear with me here...why must $\lim_{L\to\infty} \int_{-\infty}^{L} f(t')\ \text{d}t$ vanish if $\int_{-\infty}^{t} f(t')\ \text{d}t$ is square integrable? Reading the wiki page on it isn't seeming to help... – AtticusFinch95 Apr 29 '16 at 11:53
• You're welcome. My pleasure. -Mark – Mark Viola Apr 29 '16 at 13:28
• Thank you for the catch. I've edited accordingly. I also added the previously tacitly assume assumption that $f$ is continuous. Then, $\int_{-\infty}^t f(t')\,dt'$ is continuous and differentiable (this permits the integration by parts). Moreover, if $\int_{-\infty}^t f(t')\,dt'$ is square integrable and continuous, then it must vanish at $\infty$. -Mark – Mark Viola Apr 29 '16 at 13:31
• I've got it now, thank you. Ah, I was just a lowly newbie beforehand and couldn't upvote any answers, but I can now. Done. ;) – AtticusFinch95 Apr 29 '16 at 13:51
• Thank you; much appreciative! I must ask ... is Atticus Finch your real name or are you a fan of "To Kill a Mockingbird?" – Mark Viola Apr 29 '16 at 14:18