I was studying Fourier Transform; I could answer to this $$\int_{-\infty}^\infty e^{-i\omega t}dt$$ by Fourier Transform, but I have problem in $$\int_0^\infty e^{-i\omega t}dt.$$ I would be grateful if you help me to get the point of this integral.

  • $\begingroup$ So you want to integrate $$\int_0^\infty e^{-jwt}dt.$$ $\endgroup$ – GWu Apr 11 '11 at 18:05

It's the Fourier transform of the Heaviside step function (which is zero for $x<0$). It only exists in the sense of distributions. Does it help if I point you to entry number 313 in Wikipedia's table of transforms?

  • $\begingroup$ the Fourier transform actually exists for $\text{Im}\, \omega < 0$. $\endgroup$ – Fabian Apr 11 '11 at 19:28
  • $\begingroup$ @Fabian: Sure, but when one says "Fourier transform" rather than "two-sided Laplace transform" it is usually understood that $\omega$ is a real number. $\endgroup$ – Hans Lundmark Apr 11 '11 at 20:28
  • 2
    $\begingroup$ I was not aware of the difference. For me Laplace and Fourier transform is all the same just a bit rotated ;-) $\endgroup$ – Fabian Apr 11 '11 at 20:35

Assuming $\text{Im}\,\omega<0$ (for convergence), you can calculate the integral easily $$\int_0^\infty e^{-i\omega t}dt = \frac{e^{-i\omega t}}{-i \omega} \biggl|_{t= 0}^{\infty} = \frac{i}{\omega}.$$

As some people want to use the fourier transform for $\omega$ on the real line, they write $$\frac{i}{\omega - i 0^+}$$ where the $0^+$ reminds them that the original integral was only defined for $\omega$ in the lower half plane.

At some point, you might learn about distributions (which Hans Lundmark was referring to) and then you can use the Sokhatsky-theorem and rewrite it as $$\int_0^\infty e^{-i\omega t}dt = \mathcal{P}\frac{i}{\omega} - \pi \delta(\omega).$$

  • $\begingroup$ Thank you so much for your kind help.I'm thinking and studying to learn what you mentioned about Sokhatsky Theorem and distributions. $\endgroup$ – user9452 Apr 12 '11 at 4:19

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