Fourier transform and inverse transform

[babyrudin]

Homework Statement

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
\[f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.\]

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?

 

[HallsofIvy]

Homework Statement

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
\[f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.\]

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?

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)

 

[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.