# Finding $\lim \limits_{R \rightarrow \infty} \int _{\Gamma_R} \frac1{(z+i)^2 (z-i)^2}dz$

Let $\Gamma_R$ be the half circle centred at $0$ and radius $R>3$ with $Im(z) \geq 0$. Show that $$\lim \limits_{R \rightarrow \infty} \int _{\Gamma_R} \frac1{(z+i)^2 (z-i)^2}dz=0$$

Is this region even closed? And I was thinking of using the $ML$ lemma but wouldn't that show it is zero for the modulus of the integral?

Please help. The denominator was originally $(z^2 +1)^2$ but i changed it just incase we might have to use Residue theorem. Would showing $$\Bigg|\int _{\Gamma_R} \frac1{(z+i)^2 (z-i)^2} \Bigg| \leq ML$$ and then showing $ML \rightarrow 0$ answer the question?

• i'm not sure if $\Gamma$ denotes the half circle without the real line or is it included? The proposition only makes sense if not. Otherwise the value would be $\pi/2$. – tired May 27 '15 at 12:35
• I wrote it word for word. So should we assume that it is a semi circle with a horizontal line in the real line from -R to R? – cooldudeman May 27 '15 at 12:37
• Hint: $\frac{1}{|(z+i)^2(z-i)^2|} \le \frac{1}{(R-1)^4}$ on $\Gamma_R$. – achille hui May 27 '15 at 12:37
• It cannot be a semicircle with a horizontal line. Otherwise, the limit of the integral won't be zero. – achille hui May 27 '15 at 12:38
• @achillehui Can we do, since $|z|=R$ then since $$\frac1{ \big|(z^2 +1)^2 \big|} =\frac1{|z^2+1|^2} \leq \frac1{ (|z|^2 -1)^2} < \frac1{|z|^2-1}=\frac1{R^2 -1}=M$$ and $L=\pi R$. By the $ML$ lemma, $$\Bigg| \int _{\Gamma_R} f(z)dz \Bigg| \leq ML$$ and since $ML \rightarrow 0$ and $R \rightarrow \infty$, this gives us the answer to the original question???? – cooldudeman May 27 '15 at 12:44

$$\left|\int_{\Gamma_R}\frac{dz}{(z^2+1)^2}\right|\le\frac{\pi R}{R^4-1}\xrightarrow[R\to\infty]{}0$$