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Find the Fourier Transform of $e^{ixt}$, where $x$ is a real parameter, $t\in \mathbb R$. I started writing: $$\int_{-\infty}^\infty e^{ixt} e^{-i\omega t}dt$$ but I do not know how to go on!

Edward
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Mark
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2 Answers2

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$$\int_{-\infty}^{+\infty} e^{ixt} e^{-i \omega t} dt=\int_{-\infty}^{+\infty} e^{ixt-i \omega t} dt=\int_{-\infty}^{+\infty} e^{i(x- \omega )t} dt=2 \pi \delta(x- \omega)$$

EDIT: It is known that:

$$\delta(x-a)= \frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{ip(x-a)} dp$$

evinda
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    Just curious, how would you justify the value of this integral at any other point than $x = \omega$? I can't see how it converges. – Arthur Jul 20 '14 at 12:01
  • For $x=\omega$,the integral is equal to:

    $$\int_{-\infty}^{+\infty} 1 dt$$ and so,obviously,it does not converge.Also,the integral diverges for $x \neq \omega$.The fact that $||e^{i(x- \omega)t}||=1$ is enough,to conclude this.

     – evinda Jul 20 '14 at 12:55
  • That is true, and I have no trouble accepting this. However, at the other points? I cannot see the integral converging to any one value. So, at $x = \omega$, I can buy the $\delta$-evaluation. But why is it chosen to be zero everywhere else? – Arthur Jul 20 '14 at 12:58
  • We know that:

    $$\int_{-a}^a e^{it} dt=2 \sin a$$

    and the limit $a \to \infty$ does not exist.

    So,the integral does not diverge for $x \neq \omega$

     – evinda Jul 20 '14 at 14:05
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Using the inverse Fourier transform, you may write (everything in distributional sense)

$f(x)=\frac{1}{2\pi}\int_\mathbb{R}e^{ixt}\int_\mathbb{R} e^{-i\omega t} dt d\omega =\frac{1}{2\pi}\int_\mathbb{R}(\int_\mathbb{R}e^{ixt} e^{-i\omega t}dt) d\omega=\frac{1}{2\pi}\int_\mathbb{R}\delta(x-\omega)f(\omega)d\omega $

And therefore your Integral equals $2\pi \delta(x-\omega)$ in the distributional sense.

Daniel
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