My Math Forum Question about Fourier transforms

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 June 29th, 2017, 11:20 AM #1 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 Question about Fourier transforms Hi all, I came up with the following question during a calculation. Suppose I have a function $f(x)$ defined only for $x>0$. Then the Fourier transform can be written as: \begin{eqnarray} f(p)=\int _{0}^{+\infty} dx e^{ipx}f(x)=\int _{-\infty}^{+\infty} dx e^{ipx}f(x)\theta(x) \end{eqnarray} Now suppose I wanna split this Fourier transform into two components, so that: \begin{eqnarray} f(p)=f_{A}(p)+f_{0}(p) \end{eqnarray} where \begin{eqnarray} f_{A}(p)=\int_{0}^{A}dx e^{ipx}f(x) && f_{0}(x)=\int_{A}^{+\infty}dx e^{ipx}f(x) \end{eqnarray} Now, I calculate $f_{A}(p)=f(p)-f_{0}(p)$. Then the question is the following: When I Fourier transform back $f_{A}(p)$, we have: \begin{eqnarray} f(x)_{A}=\int_{-\infty}^{+\infty}(dp/2\pi)e^{-ipx}f_{A}(p) \end{eqnarray} So is the function $f_{A}(x)$ defined in the region $x\in[0,A]$ only? Thank you !
 June 29th, 2017, 12:15 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,276 Thanks: 2437 Math Focus: Mainly analysis and algebra No. Your original function was implicitly defined on $(-\infty,+\infty)$ with $f(x)=0$ for $x \lt 0$. Work it through for $$f(x)=\begin{cases}0 & (x \lt 0) \\ 1 & (x \ge 0)\end{cases}$$ Last edited by v8archie; June 29th, 2017 at 12:19 PM.
June 29th, 2017, 01:33 PM   #3
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Quote:
 Originally Posted by v8archie No. Your original function was implicitly defined on $(-\infty,+\infty)$ with $f(x)=0$ for $x \lt 0$. Work it through for $$f(x)=\begin{cases}0 & (x \lt 0) \\ 1 & (x \ge 0)\end{cases}$$
Ok, so does that mean that $f_{A}(x)$ is defined in $x\in (-\infty,+\infty)$ then?

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