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I know that when $f(x)$ is a $2 \pi$ periodic function, integrable on the circle $\mathbb{T}$, then for any $n \in \mathbb{N}$

$\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(k) e^{-ikn} dk= \hat{f}(n).$

Now, suppose that we have the same integral, but with $n= q+r/m$, with $r/m <1$ and $q \in \mathbb{N}$. i.e. $n$ is not an integer. Then, how can we define

$\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(k) e^{-ikn} dk=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(k) e^{-ik(q+r/m)} dk ?$

Is it simply equal to zero?

abc
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    Then you get the Fourier Transform. – Paul Sinclair Jan 12 '18 at 20:03
  • @PaulSinclair But I thought Fourier transform where from $-\infty$ to $\infty$. my integral is only on the circle – abc Jan 12 '18 at 20:40
  • I suppose I should have looked closer at what you are doing. You say you are integrating $k$ over "the circle" $\Bbb T$, but then $k$ appears in the exponent of $e$, which is ill-defined if $k$ is not real. What exactly do you mean by "the circle" here? – Paul Sinclair Jan 12 '18 at 22:04
  • well, we know $\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(k) e^{-i kn} dk = \hat{f}(n)$ for n $\in \mathbb{Z}$. my question is what happens if $n=q+r/m$ for $q \in \mathbb{N}$ and $r/m <1$. @PaulSinclair – abc Jan 13 '18 at 01:48

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You can define $g(k)=f(k)e^{-ikr/m}$. Your integral is then $\hat g(q)$

Ross Millikan
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  • I have seen $g(x)$ and $g'(x)$ but I have not seen $\hat g(x)$. Could you please explain to me what that is? .... I did a little bit of research and it is the fourier transform of a function $g$. Nevermind... I am not ready for that yet haha – Mr Pie Jan 13 '18 at 03:46
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    It is the same transform on $g$ as $\hat f$ is on $f$. My point is you just incorporate the $e^{-ikr/m}$ into $g$, then transform it the same way you did $f$. – Ross Millikan Jan 13 '18 at 03:50