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Let $\{a_h\}$ be a double-sided complex sequence such that $\sum_{h=-\infty}^{\infty} |a_i| <\infty$ with $a_{0}\neq0$.

Set $f(x) := \sum_{h=-\infty}^{\infty} a_h \exp(ixh)$ and assume that $f(x) >c >0$.

I wish to compute the integral \begin{eqnarray} \int _{-\pi}^{\pi} \frac{1}{ f(x)} dx. \end{eqnarray} My conjecture (or hope) is that \begin{eqnarray} \int _{-\pi}^{\pi} \frac{1}{f(x)} dx = \frac{2\pi}{a_0}. \end{eqnarray} Is this the case?

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up vote 2 down vote accepted

No, this is not the case. For instance, for $f(x)=1+\frac12\sin x$, that integral is $4\pi/\sqrt3\ne2\pi$.

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There is something that I would like to understand. At wikipedia, , you see that inversion of the Fourier series would give $\frac{1}{f(x)}=\frac{1}{a_0}-\frac{a_1}{a_0^2}e^{ix}+\ldots$. How does this cope with your counterexample and the OP hypothesis? – Jon Mar 16 '12 at 11:55
@Jon: That result is for power series with positive exponents; the Fourier series has positive and negative exponents. If you replace $\sin x$ by $\mathrm e^{\mathrm ix}$ in my example, the integral is $2\pi$, as predicted by the power series result. You can think of it as writing the denominator as $a_0+x$ and expanding around $x=0$; then if $x$ contains only positive powers, no other constant terms appear, but if it contains positive and negative powers, they do. – joriki Mar 16 '12 at 12:16
Thanks for the clarification. – Jon Mar 16 '12 at 12:53
Thank you for your help. From the arguments above, can I understand that $\int_{\pi}^{\pi} 1/f(x) dx = 2\pi/a_0$ holds if the series only has nonnegative exponents? – Yamamoto Mar 17 '12 at 12:22
@Haruo: Note that the Wikipedia article Jon quoted is about formal power series, so you might also have to think about convergence. At least formally, though, if the series has only non-negative exponents, the constant term of its reciprocal is the reciprocal of the (non-zero) constant term. Note that for Fourier series, a series with only non-negative exponents is necessarily complex. – joriki Mar 17 '12 at 18:27

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