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A function $f : \mathbb{C} \to \mathbb{C}$ is analytic in the open disk $|z|<1$. I have an integral $I = \int\limits_{C} f(z) \frac{P(z)}{Q(z}$ where $C$ is $|z| = 1$ and $P(z)$ and $Q(z)$ are polynomials. I'd like to ask how i could go about evaluating such an integral.

EDIT :

$f$ is defined and continuous on the closed unit disk.

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Is $f$ defined and continuous on the closed unit disk? –  Akhil Mathew Jun 10 '11 at 12:59
    
@Akhil Mathew : yes –  Rajesh D Jun 10 '11 at 13:01

2 Answers 2

up vote 3 down vote accepted

Apply the residue theorem to circles $\gamma_{1-\epsilon}$ of radius $1-\epsilon$. If $Q$ has no zero on the unit circle then the integral $\int_{\gamma_{1 -\epsilon}}\ldots \ $ will converge to $\int_{\gamma_1}\ldots \ $ with $\epsilon \to 0+$.

This means that $$\int_{\gamma_1} f(z){P(z)\over Q(z)}dz=\lim_{\epsilon\to 0+} \int_{\gamma_{1-\epsilon}} f(z){P(z)\over Q(z)}dz=2\pi i\ \lim_{\epsilon\to 0+} \sum_{z\in D_{1-\epsilon}}{\rm res}\Bigl(f(z){P(z)\over Q(z)}\Bigr)\ ,$$ and the latter limit is, of course, equal to $$\sum_{z\in D}{\rm res}\Bigl(f(z){P(z)\over Q(z)}\Bigr)\ ,$$ as expected.

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thank you for the answer. I am almost desperate nedd of the answer, but i am not able to understand your answer, request you to kindly elaborate a bit and also if needed suggest some quick reference books. –  Rajesh D Jun 10 '11 at 15:38
    
thank you very much for the edit. –  Rajesh D Jun 12 '11 at 4:16

The contour integral may be reduced to the integral around the singularities of the integrand. The latter coincide with the singularities of $Q(z)$ in the denominator - in other words, zeros of $Q(z)$.

The result is $2\pi i$ times the sum of the residues. The residues are the coefficients of $1/(z-z_i)$ in the Laurent expansion of the integrand around the roots $z=z_i$ of $Q(z)$. If $Q(z)$ only has simple roots, the $i$th residue is just $$\lim_{z\to z_i} f(z_i)P(z_i) (z-z_i)/Q(z_i)$$

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for residue theorem to be applicable $f(z)$ should be analytic on the contour as well, i.e, in this case $f(z)$ should be analytic on the the closed unit disk. –  Rajesh D Jun 10 '11 at 13:14

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