1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2011 at 15:50
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2011 at 16:02
  • $\begingroup$ @Antoine I am amused that pastein.com is just like mathurl... $\endgroup$
    – Igor Rivin
    Dec 14, 2011 at 16:04
  • $\begingroup$ There's a lot of sites like that, usually used for pasting code. I didn't actually know about mathurl, it's pretty neat! $\endgroup$ Dec 14, 2011 at 16:08

2 Answers 2

5
$\begingroup$

$$ \begin{aligned} &\int_0^\pi t^\beta e^{iyt}\frac{dt}{t}dt=y^{-\beta}\int_0^{\pi y} t^{\beta} e^{it}\frac{dt}{t}= \cr &=y^{-\beta}e^{i\pi\beta/2}\left[\int_0^{\pi y} t^\beta e^{-t}\frac{dt}{t}+O(t^{\beta-1})\right]= y^{-\beta}e^{i\pi\beta/2}[\Gamma(\beta)+o(1)]. \end{aligned} $$

Now just take the real part. If you want more terms, apply Laplace on the arc used for moving the interval of integration to the imaginary axis.

$\endgroup$
0
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.