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I'm reading Boas' book and there's a part where she assumes that the following integral goes to $0$ as $\rho$ tends to $\infty$. The argument is that the numerator contains $\rho$ while the denominator contains $\rho^2$. However, I'm uncomfortable with this line of reasoning since the integral has not yet been evaluated.

$$\int_0^{\pi} \frac{\rho ie^{i\theta}d\theta}{1+{\rho^2 e^{2i\theta}}}$$

Can anybody (a) guide me on how to evaluate the integral (b) tell me why the above reasoning is sound in spite of not evaluating the integral? Perhaps if I could do the first part, it would shed some light on the second. Thanks.

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The fact you want to use is that $$ \left| \int_\gamma f(z) \, dz \right| \le \int_\gamma |f(z)| \, |dz| $$ which in your case means that $$ \left| \int_0^\pi \frac{ \rho i e^{i\theta} d\theta }{ 1+\rho^2e^{2i\theta} } \right| \le \int_0^\pi \frac{ \rho d\theta }{ \rho^2 - 1 } = \frac{\rho}{\rho^2-1} \pi \xrightarrow[\rho\to\infty]{} 0$$ Notice that you never need to actually evaluate the original integral.

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$$\int_\rho^{-\rho}\frac{du}{1+u^2}=\left.\arctan u\right|_\rho^{-\rho}=\arctan(-\rho)-\arctan \rho\xrightarrow [\rho\to\infty]{}-\frac{\pi}{2}-\frac{\pi}{2}=-\pi\neq 0$$Did I miss something? – DonAntonio Jul 19 '12 at 14:02
woops! Thanks DonAntonio. Those two integrals cannot be the same. The original integral looks like $\int_\gamma \frac{dz}{1+z^2}$ where $\gamma$ is a semi-circle from $\rho$ to $-\rho$. Whereas the integral in my comment is $\int_{\gamma'} \frac{dz}{1+z^2}$ where $\gamma'$ runs along the real axis from $\rho$ to $-\rho$. You cannot deform $\gamma$ to $\gamma'$ without passing through a singularity of $\frac1{1+z^2}$. – Morgan Sherman Jul 19 '12 at 19:36

If the limit of the integrand converges uniformly, then you can take the limit under the integral sign. In your case, the limit of the integrand does converge uniformly. Taking the limit of the integrand then integrating you get the answer $0$. Or more general, you can use the Lebesgue Dominated Convergence Theorem.

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