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Mathematica fails to find a Fourier transform of exponent.

Yet according to this page

$$\mathcal{F}[e^{2\pi iat}]=\delta(t-a)$$

and via substitution,

$$\mathcal{F}[e^{at}]=\delta\left(t-\frac a{2\pi i}\right)$$

Yet this does not make much sense because inverse Fourier transform of this will give 0.

Thus my question is what one can do so to fix this and make Fourier transform of exponent useful and revertible.

Possibly some modification of Dirac Delta for complex argument? Or taking Fourier integral over complex plane? Or taking two integrals, one over reals the other over imaginary axis?

Mister Benjamin Dover
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Anixx
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  • In the substitution, is a real or complex? Convergence of the Fourier transform is guaranteed only when a is imaginary or when $|a|<1$.And if a is real, the substitution is incorrect. – Karthik Upadhya Oct 14 '14 at 06:40
  • Why inverse Fourier transform is 0 ? – Inquisitive Oct 14 '14 at 07:29
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    May be this is useful: The Fourier transform is an isomorphism on $\mathcal{S'}(\mathbb R^{n})$(=The space of tempered distributions) onto itself.(Note that Dirac delta is in $\mathcal{S'}(\mathbb R^{n})$). – Inquisitive Oct 14 '14 at 07:33
  • The Fourier transform of $f(t)=e^{at}$ does not exist for $a\in\mathbb{R}$. Things look different if the function is an exponential function only for $t>0$ (and zero for $t<0$). – Matt L. Oct 14 '14 at 09:11
  • @Inquisitive because integral of that function along the real line is 0. – Anixx Oct 14 '14 at 09:23
  • @Inquisitive this exactly follows from the first formula in the question so what? – Anixx Oct 14 '14 at 10:12

1 Answers1

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Mathematica $10.0.2.0$ can find the Fourier transform of $f(t) = e^{2\pi i at}$. enter image description here

Mathematica defines the Fourier transform as $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt $$ Now we can derive the solution shown by Mathematica and the website. \begin{align} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{2\pi iat}e^{i\omega t}dt &= \frac{1}{\sqrt{2\pi}}\lim_{x\to\infty}\int_{-x}^{x}e^{it(2\pi a+\omega)}dt\\ &= \sqrt{2\pi}\lim_{x\to\infty}\frac{\sin[(2\pi a+\omega)x]}{(2\pi a+\omega)\pi} \end{align} Let $y=2\pi a+\omega$ and $\epsilon=\frac{1}{x}$. Then \begin{align} \sqrt{2\pi}\lim_{\epsilon\to 0}\frac{\sin[y/\epsilon]}{y\pi} &= \sqrt{2\pi}\delta(y)\\ &= \sqrt{2\pi}\delta(\omega + 2\pi a) \end{align} Now, we can take the inverse Fourier transform $F(\omega) = \sqrt{2\pi}\delta[\omega - (-2\pi a)]$ so $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sqrt{2\pi}\delta[\omega - (-2\pi a)]e^{-i\omega t}d\omega = e^{2\pi iat}\tag{1} $$ which occurs by the sifting property of the Dirac Delta function.

dustin
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