# Fourier Transform on positive real line

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.

• What is your function exactly? Can't you define your function equal to zero on $]-\infty,0]$? Notice that you must have either $\int |f(x)| dx< + \infty$ to apply directly the Fourier Transform. Jan 24, 2018 at 9:46
• You can settle for $\int|f(x)|^2dx<+\infty$ too, no? The Fourier transform can be extended to $L^2$ functions by a density argument if I remember correctly. Jan 24, 2018 at 9:49
• The natural transform to apply on the half line is the Laplace Transform. Typically Fourier transform is applied on the whole line and Fourier Series to finite intervals.
– Paul
Jan 24, 2018 at 9:59
• @Netchaiev My function in the simplest case is simply $e^{-x}$ defined over $x\geq 0$. It is the probability that a droplet does not suffer a collision after it has fallen a distance $x$ (oriented downward). Since a droplet falls and does not rise the probability function is not defined for $x<0$. Assuming it to be zero for $x<0$ is physically erroneous; it is simply not defined.
– Deep
Jan 24, 2018 at 10:36
• @TonyS.F. Please see my comment above.
– Deep
Jan 24, 2018 at 10:37

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.
• +1 It seems that Cosine/Sine transforms are also defined over $(-\infty,+\infty)$ but for even (odd) functions the integral reduces to Cosine (Sine) transform over $[0,\infty)$ as per en.wikipedia.org/wiki/Sine_and_cosine_transforms.
• @Deep : The Fourier sine transform also naturally arises in connection with the spectral theory for $-\frac{d^2}{dx^2}$ on $[0,\infty)$ with $0$ endpoint condition at $x=0$. The cosine transform is associated with the same operator on $[0,\infty)$ but $0$ first derivative condition at $0$. Jan 25, 2018 at 4:40