How do I calculate the Fourier transform ($t \rightarrow \omega$) of the following:
$\exp(A\cos(\omega_0 t))$
$A$ is a real constant, and $\omega_0$ is a real and positive constant. I know that this gives a Bessel function, but how can I show this?
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1What do you mean about reputation? There's no reputation requirement to tag a question. – Jun 17 '14 at 01:52
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@user61527 Apologies – Medulla Oblongata Jun 17 '14 at 01:53
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Hello and welcome to math.stackexchange. Does the frequency variable $\omega$ really appear in the function that you want to transform or is that perhaps a typo? – Hans Engler Jun 17 '14 at 01:57
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@HansEngler No, not a typo. The $\omega$ is a constant. I'll rename it $\omega_0$ if you prefer. – Medulla Oblongata Jun 17 '14 at 02:00
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You could try to show that the Fourier transform satisfies a suitable differential equation. Of course there is the small issue of convergence of the Fourier integrals. – Hans Engler Jun 17 '14 at 02:06
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The Fourier transform of this function does not exist in the usual sense since this function is not in any $L^1$. It has to be evaluated distributionally. Why do you think it gives a Bessel function as an output? – Cameron Williams Jun 17 '14 at 02:09
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According to Abramowitz and Stegun's Handbook of Mathematical Functions, it is a Bessel function... – Medulla Oblongata Jun 17 '14 at 02:10
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Can you please provide the claim by the authors? I am very skeptical of this. – Cameron Williams Jun 17 '14 at 02:12
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The modified Bessel Function of second kind $K_z(a)$ can be expressed as a Fourier transform see here:
$$K_{z}(a)=\frac{1}{2}\int_{-\infty}^{\infty}\exp(-a\cosh t)\cosh(zt){\rm d}t=K_{-z}(a)$$
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I would say this is related for sure, but this integral is at least well-posed. OP's is not. The function OP is interested in is very, very badly behaved with regards to the Fourier transform. Unless of course $\cosh$ is meant instead of $\cos$... – Cameron Williams Jun 17 '14 at 02:32
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Only to the extent that $cos$ itself is "badly behaved" with regards to the Fourier transform. I don't think just being a Dirac comb is sufficient for that... – linkhyrule5 Feb 26 '22 at 01:49