Let $R \in \mathbb C$. Is it possible to show $$ \lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta $$ using the dominated convergence theorem?
Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
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complex-analysis
measure-theory
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1What do you call $\sqrt{1+R^{-2}\mathrm e^{-2i\theta}}$? – 2012-09-23
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0@did This is the radius of the complex number $1+iR^{-1}e^{-i\theta}$. – 2012-09-23
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0@did So $\|1+iR^{-1}e{-i\theta}\|^2=1+R^{-2}e^{-2i\theta} \ge 1$, and DCT follows. Is it correct? – 2012-09-23
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1No, the radius of the complex number $1+iR^{-1}e^{-i\theta}$ is $\sqrt{1+|R|^{-2}-2\Im(e^{-i\theta}R^{-1})}$. – 2012-09-23
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0@did I see your point. How do I get a lower bound for this radius? – 2012-09-23
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1This is $\geqslant\sqrt{1+|R|^{-2}-2|R|^{-1}}=1-|R|^{-1}$ if $|R|\geqslant1$. – 2012-09-23
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0@did We showed $\sqrt{1+|R|^{-2}-2\Im(R^{-1}e^{-i\theta})} \ge 1-|R|^{-1}$. But I need a lower bound for $\sqrt{1+R^{-2}e^{-2i\theta}}$ to apply the DCT. What is the link between these? – 2012-09-25
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1Once again, $\sqrt{1+R^{-2}e^{-2i\theta}}$ has no (canonical) definition. The modulus of $1+iR^{-1}e^{-i\theta}$ is given by the formula I wrote, **not by** $\sqrt{1+R^{-2}e^{-2i\theta}}$, **which has no meaning**. And the best uniform lower bound for the modulus of $1+iR^{-1}e^{-i\theta}$ **is the one I gave you**. (Believe it or not, my comments are not composed randomly...) – 2012-09-25
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0@did I believe you are pretty strong, finding this lower bound on the spot. Any idea for the original question? – 2012-09-25
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1The modulus is between $1-|R|^{-1}$ and $1+|R|^{-1}$, uniformly on $\theta$ hence yes, the limit of the integral exists and it is $2\pi i$. – 2012-09-25