If $f(x)$ is a periodic signal with a period $A$ then the Fourier transform of the signal $F(k)$ is zero, unless $k A$= $2 \pi n$ where $n$ is an integer. How can there Be a Fourier transform of a function which do not converge as $x$ goes to (-$\infty$ to $\infty$)?
besides even if the integration carried within the period $A$ there is no information given whether f(x) is odd or even function to turn to the Classical definition of Fourier Transform. How in general the Fourier Transform for a periodic signal exist as long as $k$= $2pi$ $n$
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1$\begingroup$ Well, the key-word here is Fourier series. The Fourier integral is an integral, and that's required when your function is defined everywhere and isn't periodic. But the analog of this for periodic functions is the Fourier series, which is an infinite sum over the allowed $k$'s. I'm not sure exactly what your question really is, without some examples, but perhaps looking up information about Fourier series can help you here. $\endgroup$– marchCommented Jun 5 at 15:11
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$\begingroup$ The Fourier transform integral will give you a series of Dirac delta functions, so the integral does converge but only in the sense of a distribution. In other words you need to consider the machinery of test fuctions to evaluate it rigorously. $\endgroup$– mike stoneCommented Jun 5 at 16:21
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