I need to prove:

$$\lim_{R\to +\infty} \left|\int_0^\pi \frac{e^{iaR(\cos\theta+i\sin\theta)}}{(1+R^2e^{2i\theta})^2}iRe^{i\theta} d\theta\right| =0$$

Could someone give me some pointers? A solutions manual states: $$\left|\int_0^\pi \frac{e^{iaR(\cos\theta+i\sin\theta)}}{(1+R^2e^{2i\theta})^2}iRe^{i\theta} d\theta\right| \leqslant \frac{\pi R}{(R^2-1)^2}$$

I know how to clean op the numerator, but how does one prove $|(1+R^2e^{2i\theta})^2| \geqslant (R^2-1)^2$.

I have tried: $|(1+R^2e^{2i\theta})^2|=|1+2R^2e^{2i\theta}+R^4e^{4i\theta}|\geqslant\left|1-2R^2-R^4\right|$ because of the reverse triangle inequality, but then...


1 Answer 1


It's enough to prove that $|1 + R^2e^{2i \theta}| \geq R^2 - 1$. This comes easily from the reverse triangle inequality: $$ |1 + R^2 e^{2i\theta}| \geq |R^2 e^{2i\theta}| - |1| = R^2 - 1.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .