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babyrudin
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Homework Statement
Let f(x) be an integrable complex-valued function on [tex]\mathbb{R}[/tex]. We define the Fourier transform [tex]\phi=\mathcal{F}f[/tex] by
[tex]\[\phi(t)=\int_{\infty}^{\infty} e^{ixt} f(x) dx.\][/tex]
Show that if [tex]f[/tex] is continuous and if [tex]$\phi$[/tex] is integrable, then
[tex]\[f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.\][/tex]
The Attempt at a Solution
So far I have
[tex]\[\int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt = \lim_{a \to \infty} \int_{-a}^{a} e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-a}^a e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-\infty}^\infty \frac{2 \sin((y-x)a)}{y-x} f(y) dy.\][/tex]
What to do next?
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