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I'm currently self-studying complex analysis and I'd like to analytically show that $$\lim\limits_{R\to\infty}\int_{\gamma_R} \frac{\ln\left(z+i\right)}{z^2+1}\ \mathrm dz=0$$ Where $$ \gamma_R=\left\{Re^{it}:t\in [0, \pi]\right\}$$ After setting $z=Re^{it}$, using the triangle inequality and applying the estimation lemma, I end up with $$\int_0^{\pi} \left|\frac{\ln\left(Re^{it}+i\right)}{\left(Re^{it}\right)^2+1}\right|\left|iRe^{it}\right|\ \mathrm dt\leq\frac{\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}\int_0^{\pi} \left|iRe^{it}\right|\ \mathrm dt$$ $$\int_0^{\pi} \left|\frac{\ln\left(Re^{it}+i\right)}{\left(Re^{it}\right)^2+1}\right|\left|iRe^{it}\right|\ \mathrm dt\leq\frac{\pi R\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}$$ I'm intuitively aware that $$\lim\limits_{R\to\infty}\frac{\pi R\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}=0$$ My question is how do I proceed in analytically showing that the aforementioned limit is indeed zero? My trouble seems to come from a lack of understanding of how the modulus of a complex log works. Any additional insight on this is very appreciated and thank you for reading my post.

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    $\begingroup$ If $z=re^{i\theta}$, then $|\log z|=\sqrt{(\log r)^2 +{\theta_k}^2}$, where $\theta_k=\theta+2\pi k$ is the appropriate argument for the chosen branch of the logarithm. Also, note that you can't bring the variable of integration outside of the integral. You need to estimate the value, as you indicate. $\endgroup$ – MPW Mar 24 '16 at 2:50
  • $\begingroup$ +1, That definition helped me a lot. Also thanks for catching my silly mistake. Errors like that eat away at my soul. $\endgroup$ – k170 Mar 24 '16 at 16:55
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To be precise, let's cut the plane at the branch point $z=-i$ with a straight line along the negative imaginary axis to $z=-i\infty$. Then, on $\gamma_R$, we have

$$0<\arctan(1/R)\le \arg(z+i)\le \pi -\arctan(1/R)<\pi$$

Therefore, on $\gamma_R$, the magnitude of the complex logarithm is bounded by

$$\begin{align}|\log(Re^{i\phi}+i)|&\le \sqrt{\log^2\left(\sqrt{R^2+2R\sin(\phi)+1}\right)+\pi^2}\\\\ &\le \sqrt{\log^2\left(R+1\right)+\pi^2} \end{align}$$

Finally, we can bound the integral of interest by

$$\begin{align} \left|\int_{\gamma_R}\frac{\log(z+i)}{z^2+1}\,dz\right|&=\left|\int_0^\pi \frac{\log(Re^{i\phi}+i)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\,d\phi\right|\\\\ &\le \int_0^\pi \left|\frac{\log(Re^{i\phi}+i)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\right|\,d\phi\\\\ &\le \int_0^\pi \frac{|\log(Re^{i\phi}+i)|}{|R^2e^{i2\phi}+1|}|iRe^{i\phi}|\,d\phi\\\\ &=\int_0^\pi \frac{|\log(Re^{i\phi}+i)|}{|R^2e^{i2\phi}+1|} R\,d\phi\\\\ &\le \frac{\pi\,R\,\sqrt{\log^2\left(R+1\right)+\pi^2} }{R^2-1}\\\\ &\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$

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    $\begingroup$ Thank you for taking the time to answer my question. I've been studying the first two expressions in your post and I have some questions. Just for clarity, am I correct to say that the basic strategy to bound the modulus of a complex log is to first find an upper bound to the argument for the chosen branch? Also I noticed that in the following expression $$|\log(Re^{i\phi}+i)|\le \log^2\left(\sqrt{R^2+2R\sin(\phi)+1}\right)+\pi^2$$ The right hand side isn't under a square root. Shouldn't it be as follows?$$|\log(Re^{i\phi}+i)|\le \sqrt{\log^2\left(\sqrt{R^2+2R\sin(\phi)+1}\right)+\pi^2}$$ $\endgroup$ – k170 Mar 24 '16 at 16:44
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    $\begingroup$ Note that you have made an error similar to OP's original error. You can't use the variable of integration outside of the integral, you can only estimate the values by constants. The estimate is supposed to be something like $\left|\int_If\right| \leq |I|\cdot L$, where $\sup_I |f|\leq L$, right? $\endgroup$ – MPW Mar 24 '16 at 17:03
  • $\begingroup$ @MPW. Yes, thanks for the catch. +1 I inadvertently omitted the supremum over $\phi$. I've edited accordingly. -Mark $\endgroup$ – Mark Viola Mar 24 '16 at 17:35
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    $\begingroup$ @k170 Yes, you are correct on the second issue. The left-hand side was supposed to be the square of the magnitude. I've edited to correct. On the first issue, we are seeking the supremum ($0\le \phi\le \pi$) of the logarithm. In general, $|\log(z)|=\sqrt{\log^2(|z|)+(\arg(z)+2n\pi)^2}$. So, over an integration path on which both $|z|$ and $\arg(z)$ change, we can bound the magnitude of the logarithm by choosing the suprema for both $|z|$ and $\arg(z)$. That is the case here. -Mark $\endgroup$ – Mark Viola Mar 24 '16 at 17:43
  • $\begingroup$ @Dr.MV, You just made that very clear for me. Thanks again for your time. $\endgroup$ – k170 Mar 24 '16 at 17:47

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