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What is the Fourier sine and cosine transform of $f(x)=1$? I have seen some sources refer to the transform of $f=1$ involving the Dirac Delta function, but this goes against the integral definition for the Fourier sine transform, for example, since

$$\int_0^\infty f(x)\sin(x t)d x,$$

diverges when $f=1$ doesn't it?

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  • $\begingroup$ Are you asking about the transform of the function $f(x) = 1$, or the dirac delta function? You seem to be confusing the two. $\endgroup$ – Calvin Lin Jun 26 '13 at 7:24
  • $\begingroup$ Yes, the transform of $f(x)=1$. $\endgroup$ – Pixel Jun 26 '13 at 10:34
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You can extend the Fourier transform to distributions like the Dirac Delta function. Taking the Fourier transform of $\delta(t)$ gives

$$\int_{-\infty}^{\infty}\delta(t)e^{-i\omega t}\;dt=1$$

because

$$\int_{-\infty}^{\infty}\delta(t)f(t)\;dt=f(0)$$

If you apply the (inverse) Fourier transform to 1 you get

$$\mathcal{F}^{-1}1=\text{p.v.}\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\;d\omega=\frac{1}{\pi}\int_{0}^{\infty}\cos\omega t\;d\omega$$

Of course this is a divergent integral, but if it is used in a convolution integral it does have a meaning and it is useful to define

$$\delta(t)=\text{p.v.}\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\;d\omega=\frac{1}{\pi}\int_{0}^{\infty}\cos\omega t\;d\omega$$

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