asdf1
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why doesn't e^x have Fourier cosine/sine transforms?
Fourier cosine/sine transforms
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It is too singular as x-->infty.
asdf1
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could you please explain the meaning of singular?
JohanL
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f(x) must vanish as x goes to infinity in order for the the Fourier transform of f(x) to exist
asdf1
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i see! thank you very much!
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