Math for physics - Fourier inverse transform

Question

I have basic problem. I am not able to find (in Google) and derive on myself the inverse Fourier transform integral formula. $$Fourier\space defined:$$ $$F(\omega)=\int\limits_{-\infty}^{+\infty} f(t) \cdot e^{- i \omega t} dt.$$ And now I would like to find general inverse transform of it. Could anyone help me?

The goal is: $$f(t) = \frac{1}{2\pi} \int \limits_{-\infty}^{+\infty}F(\omega)\cdot e^{ i \omega t} d\omega.$$

Answer 1

A rigorous proof can be found in most textbooks on real analysis geared toward mathematicians. Here is a heuristic derivation.\begin{align} \int_{-\infty}^{\infty}F(\omega)\exp(\mathrm{i}\omega t)d\omega&=\int_{-\infty}^\infty\left[\int_{-\infty}^\infty f(t')\exp(-\mathrm{i}\omega t')dt'\right]\exp(\mathrm{i}\omega t)d\omega\\ &=\int_{-\infty}^\infty f(t')\left[\int_{-\infty}^\infty\exp\mathrm{i}\omega(t-t')d\omega\right]dt'\\ &=2\pi\int_{-\infty}^\infty f(t')\delta(t-t')dt'\\ &=2\pi f(t) \end{align} where I used $$\delta(t-t')=\frac{1}{2\pi}\int_{-\infty}^\infty\exp\mathrm{i}\omega(t-t')d\omega.$$ Hence $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)\exp(\mathrm{i}\omega t)d\omega.$$

Answer 2

Multiply both sides of your equation for the Fourier transform with $e^{i\omega t'}$, integrate both sides with $d\omega$ and use the defining property of the delta function on the RHS. Explicitly, $$\int d\omega e^{i\omega t'}F(\omega)=\int dt\int d\omega f(t)e^{-i\omega t}e^{i\omega t'}$$

$$\implies \int d\omega e^{i\omega t'}F(\omega)=\int dt\ f(t)\int d\omega e^{-i\omega(t-t')}=\int dt f(t)\delta(t-t')$$ $$\implies \int d\omega e^{i\omega t'}F(\omega)=f(t')$$

Note that the factors of $2\pi$ are convention dependent(some will split it evenly as $1/\sqrt{2\pi}$ between both the fourier and inverse transform) so I've ignored them here, you can put them back in by properly normalising the $\delta$-function.

EDIT: Why we need the $\delta$ function-

The fourier transform is essentially a way to expand an arbitrary function $F(\omega)$ (which can be thought of as a vector in an infinite dimensional vector space) in the basis of $e^{i\omega t}$(often called the "plane-wave basis"). The analogy is, suppose you had a finite dimensional vectors, which has components $F_i$ in some given basis $\{\hat{e_i}\}$. You can thus expand $$\vec{F}=\sum_i F_i \hat{e_i}\sim \int_t f(t)e^{i\omega t}$$

and thus the $f(t)$ are to be seen as 'components' of $f(\omega)$ in the $e^{i\omega t}$ basis. Now, in the finite dimensional case, if you were to find the components $F_i$, you'd need the inner product $$F_i=\vec{F}\cdot \hat{e_i}$$

Now, the inner product on the vector space(if it exists) is completely specified by writing down the inner product between all basis vectors, i.e. it suffices to know $\langle e_i|e_j\rangle=\hat{e_i}\cdot\hat{e_j}\forall i,j$. Thus, in our $F(\omega)$ case, we need to know the inner product for the basis $e^{i\omega t}$, and this is precisely what $\delta$ function is-

$$\langle e^{i\omega t}|e^{i\omega t'}\rangle\equiv \int d\omega (e^{i\omega t})^* e^{i\omega t'}=\int d\omega e^{i\omega(t'-t)}\equiv\delta(t'-t)$$

by definition. Thus,

to find the fourier transform is to find the components in the plane wave basis, and that is found by defining the inner product of the basis vectors, and this is a delta function

Answer 3

Note: $\int := \int_{-\infty}^{\infty}$

Let $f \in L^1(\mathbb{R})$ so that $\hat{f}(\xi) = \int f(x) \, e^{-ix\xi} \, dx$ is well-defined. Then $$|\hat{f}(\xi)| = |\int f(x) \, e^{-ix\xi} \, dx| \leq \int |f(x)| \, dx < \infty$$ so $\hat{f} \in L^\infty(\mathbb{R})$, but it's not certain that $\hat{f} \in L^1(\mathbb{R})$ so that $F(y) = \int \hat{f}(\xi) \, e^{iy\xi} \, d\xi$ is well-defined. Therefore, take $\alpha>0$ and multiply $\hat{f}(\xi)$ with $e^{-\alpha\xi^2}$ to get something in $L^1(\mathbb{R})$. We will later let $\alpha\to 0$. Now, $$ \int e^{-\alpha\xi^2} \hat{f}(\xi) \, e^{iy\xi} \, d\xi = \int e^{-\alpha\xi^2} \left( \int f(x) \, e^{-ix\xi} \, dx \right) \, e^{iy\xi} \, d\xi = \int f(x) \, \left( \int e^{-\alpha\xi^2} e^{i(y-x)\xi} \, d\xi \right) \, dx $$ by changing order of integration. Here, $$ \int e^{-\alpha\xi^2} e^{i(y-x)\xi} \, d\xi = \frac{2\pi}{\sqrt{4\pi\alpha}} e^{-(x-y)^2/(4\alpha)} $$ so the integral becomes $$ \int f(x) \, \frac{2\pi}{\sqrt{4\pi\alpha}} e^{-(x-y)^2/(4\alpha)} \, dx = \{ x = y+z \} = \int f(y+z) \, \frac{2\pi}{\sqrt{4\pi\alpha}} e^{-z^2/(4\alpha)} \, dz = \{ u=z/\sqrt{4\alpha} \} = \frac{2\pi}{\sqrt{4\pi\alpha}} \int f(y+\sqrt{4\alpha}u) \, e^{-u^2} \, \sqrt{4\alpha}\,du = \frac{2\pi}{\sqrt{\pi}} \int f(y+\sqrt{4\alpha}u) \, e^{-u^2} \, du $$

Thus, $$ \int e^{-\alpha\xi^2} \hat{f}(\xi) \, e^{iy\xi} \, d\xi = \frac{2\pi}{\sqrt{\pi}} \int f(y+\sqrt{4\alpha}u) \, e^{-u^2} \, du $$ Letting $\alpha\to 0$ we get $$ \lim_{\alpha\to 0} \int e^{-\alpha\xi^2} \hat{f}(\xi) \, e^{iy\xi} \, d\xi = \lim_{\alpha\to 0} \frac{2\pi}{\sqrt{\pi}} \int f(y+\sqrt{4\alpha}u) \, e^{-u^2} \, du = \frac{2\pi}{\sqrt{\pi}} \int f(y) \, e^{-u^2} \, du = \frac{2\pi}{\sqrt{\pi}} \int e^{-u^2} \, du \, f(y) = 2\pi\, f(y) $$ If $\hat{f} \in L^1(\mathbb{R})$ this implies $$ \int \hat{f}(\xi) \, e^{iy\xi} \, d\xi = 2\pi\, f(y). $$