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I got confused with the following question when studying Fourier analysis:

If there's a function $f\in L^2(\mathbb{R})$, with the Riemann integral $\int f(t) e^{-2 \pi it \gamma} dt \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ is it true that $\int f(t) e^{-2 \pi it \gamma} dt$ is in fact the Fourier transform of $f$ defined rigorously in $L^2$ space? (i.e. the limit of a sequence of Fourier transform of functions in $L^1(\mathbb{R}) \cap L^2(\mathbb{R}).$) Thanks for any help.

user26857
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qwe0912
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  • Saying $\int f(t) e^{-2 \pi it \gamma} dt \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ makes no sense - you meant $f(t) e^{-2 \pi it \gamma} \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$. That says that $f$ is in $L^1\cap L^2$. – David C. Ullrich Sep 12 '15 at 18:32
  • I mean $g(\gamma)=\int f(t) e^{-2 \pi i t \gamma}dt$ is in $L^1(\mathbb{R}) \cap L^2(\mathbb{R})$. – qwe0912 Sep 12 '15 at 18:46
  • Assuming just $f\in L^2$ that integral doesn't exist. – David C. Ullrich Sep 12 '15 at 19:00
  • Let $f=\frac{sin(2 \pi t)}{\pi t} \in L^2$ not in $L^1$. Then $\int f(t)e^{-2 \pi i t \gamma}dt =\chi_{[-1,1]}$ is in $L^1(\mathbb{R}) \cap L^2(\mathbb{R}).$ – qwe0912 Sep 12 '15 at 19:28
  • The integral does not exist as a Lebesgue integral. If you mean some other sort of integral you need to say what you mean. The only interpretations I can think of make the question trivial, one way or another. – David C. Ullrich Sep 12 '15 at 19:53
  • Thank you for pointing it out. I guess this is why I got confused. I did the integral above by contour integral using residue theorem. I think it can also be computed by calculus, which means Riemann integral. Does the result coincide with its $L^2$-Fourier transform? – qwe0912 Sep 12 '15 at 20:16
  • Hint: What you actually calculated was $\lim_{A\to\infty}\int_{-A}^A$. – David C. Ullrich Sep 12 '15 at 20:31
  • For $f \in L^{2}$, the c.p.v. integral $\lim_{R\rightarrow\infty}\int_{-R}^{R}e^{-2\pi i t\gamma}f(t)dt$ converges in $L^{2}$ because the Fourier transform extends to a unitary map on $L^{2}$. – Disintegrating By Parts Sep 14 '15 at 21:19

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