Fourier transform and inverse transform

1. Oct 30, 2007

babyrudin

1. The problem statement, all variables and given/known data

Let f(x) be an integrable complex-valued function on $$\mathbb{R}$$. We define the Fourier transform $$\phi=\mathcal{F}f$$ by
$$$\phi(t)=\int_{\infty}^{\infty} e^{ixt} f(x) dx.$$$

Show that if $$f$$ is continuous and if $$\phi$$ is integrable, then
$$$f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.$$$

3. The attempt at a solution

So far I have
$$$\int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt = \lim_{a \to \infty} \int_{-a}^{a} e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-a}^a e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-\infty}^\infty \frac{2 \sin((y-x)a)}{y-x} f(y) dy.$$$

What to do next?

Last edited: Oct 30, 2007
2. Oct 31, 2007

HallsofIvy

Staff Emeritus
It's not a matter of "what to do next?" but "Why in the world are you doing that?". For one thing, there is no integral from 0 to $\infty$ in what you had before, where are you getting that? And even when you have two integrals, you still only have one variable: dt. You want to show that, for any continuous f (in which case $\phi$ must be integrable)
$$\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ixt} \int_{-\infty}^\infty e^{iut}f(u)dudt= f(x)$$

3. Nov 17, 2009

Arturgower

What your coarse probably wants from you is to use the following,

$$\delta(t - a) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i \omega (t -a) } d\omega.$$
If you want a full but rather simple proof, for a graduate student with a mathematical focus, of this identity I can send it to you.