# Fourier transform of an time integral

Discussion in 'Homework Help' started by epsilonjon, May 20, 2012.

1. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Question:

Derive the relationship

$\int^t_{- \infty} f(\tau) d \tau \Leftrightarrow \frac{F(\omega)}{j \omega} + \pi F(0) \delta (\omega)$

(where $\Leftrightarrow$ means "Fourier transforms into").

Attempt:

I have already proved the relationship

$\frac{dg(t)}{dt} \Leftrightarrow j \omega G( \omega)$

so define $h(t) = \frac{dg(t)}{dt}$. Then we have

$\int^t_{- \infty} h(\tau) d \tau = \int^t_{- \infty} \frac{dg}{d \tau} d \tau = [g(\tau)]^t_{- \infty} = g(t) - g(-\infty)$

so

$g(t) = \int ^t _{- \infty} h(\tau) d \tau + g(- \infty)$

Using the Fourier differentiation relationship above we get

$\mathcal{F}[h(t)] = j \omega \mathcal{F}[g(t)] = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau + g(- \infty) \right ] = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] + j \omega \mathcal{F}[g(- \infty)]$
$= j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] + j \omega 2 \pi g(-\infty) \delta(\omega)$

so

$\mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] = \frac{\mathcal{F}[h(t)]}{j \omega} - 2 \pi g(-\infty) \delta(\omega)$

I'm not sure if i've done something wrong here or if somehow this is equivalent to the correct relationship? Could someone help please

Thanks!
Jon.

Last edited: May 21, 2012