Calculation of the $\int_{-\infty}^{\infty} \frac{dx}{1+x^2}$ using contour integration I am studying complex variables. In Gamelin I found the following proof for $\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$ using contour integration here:



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*I have some basic questions, about the integral on the left in the second line being the sum of the two integrals on the right, i.e.,


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*How is the $\int_{\partial D_{R}} \frac{dZ}{1+z^2} = \int_{-R}^{R} \frac{dx}{1+x^2} + \int_{T_{R}} \frac{dz}{1+z^2}$? The area under the first integral on the right already includes the area enclosed? I am a bit confused.



*Why is he using this choice of contour? What if I choose a circle? I know if the circle includes both singularities it will go to zero. But I am not very clear.


*How is the ML estimate $\leq \frac{1}{R^2 -1}$? I expect it to be $\frac{1}{1+R^2}$
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*Firstly, please note the contour integrals are evaluated on curves, not 'surfaces'; i.e., we are not evaluating the integral on the semi-circular region (described in the excerpt from Gamelin as $D_R$), but on the curve bounding the region ($\partial D_R$), going in the anti-clockwise direction. $$\int_{\partial D_R}{\frac{dz}{1+z^2}}=\int_{-R}^{R}{\frac{dx}{1+x^2}}+\int_{\Gamma_R}{\frac{dz}{1+z^2}}$$ is simply breaking up the curve into two parts -

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*the straight line from $-R$ to $R$.

*the semi-circle $\Gamma_R$.

and evaluating the integral on the two separate parts.


*You can very well choose a circle $\mathbb{T}_R$ of radius $R$ instead. However, integrating on the circle won't tell you anything about the value of the integral on the real line, simply because the real line isn't part of your chosen curve (or contour) - the circle.
Also, as you've said - the integral on this contour, the circle $\mathbb{T}_R$ (not on the open disk of radius R - $\mathbb{D}_R$) will be zero. Can you prove that without using residues? (hint: use the symmetry of the contour)
Nonetheless, you can choose other families of contours that include at least part of the real line in each contour; and try to evaluate the integral. It is just so that this contour makes the ML estimate easier. You can also try integrating anti-clockwise on a square contour with vertices at $\{-R,R,R+2iR,-R+2iR\}$; and check your answer.
Working Advice: As far as getting ML-estimates is concerned, it is usually easier to work with square or rectangular contours for exponential, sin, or cosine functions; and with circular, semicircular, and similar radial contours for polynomial functions.


*You have gotten the inequality reversed. The triangle inequality says $$\lvert\lvert z_1\rvert-\lvert z_2\rvert\rvert\leqslant \lvert z_1+z_2\rvert\leqslant \lvert z_1\rvert+\lvert z_2\rvert$$ Putting $z_1=z^2; z_2 = 1$, you get $\lvert\lvert z\rvert^2-1\rvert\leqslant \lvert z^2+1\rvert\leqslant \lvert z^2\rvert+1$; thence, for $R>1$; $$\frac{1}{R^2-1}\geqslant \frac{1}{\lvert z^2+1\rvert}\geqslant \frac{1}{R^2+1}$$ Nonetheless, as far as this problem is concerned - this is a rather minor detail; as the main point of the ML-estimates is to estimate the order of magnitude of the integral - which, here, is $\sim \frac{1}{R} $.
