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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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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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    Thanks for detailed comments. My conjecture dose not always hold even for a Fourier series with only non-negative exponents. For specific $f$ with non-negative exponents, I have to be care about the convergence of the expansion of $1/f(x)$.2012-03-17