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