# Asymptotics of Fourier coefficients of power-type functions

I would like to understand the asymptotic behaviour of the Fourier coefficients of power type functions $f(t) = |t|^{-\alpha} 1_{[-\pi, \pi]} \qquad 0 < \alpha<1.$ I suppose this is a classic result that I am supposed to know which can be found in many books, but I do not know where to start reading. Can you give me a hint please?

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Just did some crude numerics on that for fun, result is that $\hat f(n) \approx n^{-(1 - alpha)}$. Code (python/numpy) is at pastebin.com/rL5QNMnv if you want to take a look. I'm interested in the proof. –  Antoine Levitt Dec 14 '11 at 15:50
Actually, this asymptotic behaviour is quite easy to prove using a simple change of variable. It doesn't give the full asymptotics and constants though, see Igor Rivin's answer below for that. –  Antoine Levitt Dec 14 '11 at 16:02
@Antoine I am amused that pastein.com is just like mathurl... –  Igor Rivin Dec 14 '11 at 16:04
There's a lot of sites like that, usually used for pasting code. I didn't actually know about mathurl, it's pretty neat! –  Antoine Levitt Dec 14 '11 at 16:08

\begin{aligned} &\int_0^\pi t^\beta e^{iyt}\frac{dt}{t}dt=y^{-\beta}\int_0^{\pi y} t^{\beta} e^{it}\frac{dt}{t}= \cr &=y^{-\beta}e^{i\pi\beta/2}\left[\int_0^{\pi y} t^\beta e^{-t}\frac{dt}{t}+O(t^{\beta-1})\right]= y^{-\beta}e^{i\pi\beta/2}[\Gamma(\beta)+o(1)]. \end{aligned}
This is a cosine transform of $t^{-a} \mathbb{1}[0, 1],$ which can be evaluated explicitly using the trusty mathematica, which gives: $\frac{t^{a+1} \, _1F_2\left(\frac{a}{2}+\frac{1}{2};\frac{1}{2},\frac{a}{2}+\frac{3}{2};-\frac{1}{4} k^2 t^2\right)}{a+1}$ If you want the asymptotic of the above expression in $k$ (the transform variable), you can use the mathematica command Series[your_favorite_expression, {k, Infinity, 10}] (10 gives you the first ten terms in the power series, feel free to use your favorite integer). If you use 1 instead of 10 (for ease of typesetting), you get this. (sorry, easier to use mathurl than do line breaks by hand).