Suppose $f(x)$ is defined only for $x\geq 0$. Is it correct to apply Fourier Transform operator to $f(x)$? I ask because integral of the Fourier Transform operator runs from $-\infty$ to $\infty$, but $f(x)$ is not defined for $x<0$.
P.S. Searching the net reveals examples where Fourier Transform operator is applied to partial differential equations defined over a semi-infinite domain, but that doesn't answer my question (or I don't know how it answers my question). Also I am an engineer and not a mathematician.
You can also deal with the Fourier transform on $[0,\infty)$ through cosine and or sine transforms $$ \int_{0}^{\infty}f(t)\cos(st)dt,\;\; \int_{0}^{\infty}f(t)\sin(st)dt. $$ Both of these have corresponding inverses: $$ \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(t)\cos(st)dt\right) \cos(sx)ds \sim f(x), \\ \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(t)\sin(st)dt\right)\sin(sx)ds \sim f(x). $$ Obviously there are problems with pointwise convergence at $x=0$ for the Fourier sine transform and its inverse sine transform. There are also issues for the cosine transform at $x=0$. But these are valid transforms in the $L^2$ sense, and the transforms are their own inverses, which is nice. There's also a Parseval $L^2$ identity for each.