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June 29th, 2017, 11:20 AM   #1
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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 !
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June 29th, 2017, 12:15 PM   #2
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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.
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June 29th, 2017, 01:33 PM   #3
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Quote:
Originally Posted by v8archie View Post
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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