# the integral of the inverse of a Fourier series

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 $4\pi/\sqrt3\ne2\pi$.
There is something that I would like to understand. At wikipedia, en.wikipedia.org/wiki/Formal_power_series#Inverting_series , 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
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