# how to evaluate this integral by considering $\oint_{C_{(R)}} \frac{1}{z^{2}+1}$

Consider the integral $I=\int_{-\infty}^{\infty} \frac{1}{x^{2}+1}\, dx$. Show how to evaluate this integral by considering $\oint_{C_{(R)}} \frac{1}{z^{2}+1}, dz$ where $C_{R}$ is the closed semicircle in the upper half plane with endpoints at $(-R, 0)$ and $(R, 0)$ plus the $x$ axis.

I use $\frac{1}{z^{2}+1}=-\frac{1}{2i}\left[\frac{1}{z+i}-\frac{1}{z-i}\right]$ and I must prove without using the residue theorem the integral along the open semicircle in the upper half plane vanishes as $R\rightarrow \infty$

Could someone help me through this problem?

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Use $$\left|\int_\gamma \text{blah}(z)\,dz\right|\le \int_a^b|\text{blah}(\gamma(t))\cdot \gamma\,'(t)|\,dt$$ Put a bound on the integrand so that you can derive a bound on the integral. –  anon May 10 '12 at 13:12
$$\left|\frac{1}{1+z^2}\right| \le \frac{1}{|z|^2-1}$$ if $|z| > 1$. –  mrf May 10 '12 at 15:56

Using the estimation lemma, let $$\gamma_R:=\{z\in\Bbb C\;:\;|z|=R\,,\,\Im(z)\geq 0\}\Longrightarrow \left|\oint_{\gamma_R}\frac{dz}{z^2+1}\right|\leq \sup_{z\in\gamma_R}\frac{1}{|z^2+1|}\,R\pi\leq$$ $$\leq\frac{\pi R}{R^2-1}\xrightarrow [R\to\infty]{}0$$

Also Puting $\,z=Re^{it}\,\,,\,0\leq t\leq \pi\,\Longrightarrow dz=Rie^{it}dt\,$ , so:

$$\left|\oint_{\gamma_R}\frac{dz}{z^2+1}\right|=\left|\int_0^\pi\frac{Rie^{it}dt}{1+(Re^{it})^2}\right|\leq \pi\frac{R}{R^2-1}\xrightarrow [R\to\infty]{} 0$$

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For completion's sake we will examine the function $f(z) = \frac{1}{z^2 + 1}$ parametrized by $z = z_0 + Re^{i \theta}$. If we center the contour at $z_0 = 0$ then the expansion $z^2 = R^2 e^{2it}$ so $f(z) = \frac{1}{R^2 e^{2it} + 1}$.

Given the line integral: $$\oint_{C}{f(z)\ \mathrm{d}z} = \int_a^{b}{f(C(\theta)) C'(\theta)}\ \mathrm{d}\theta$$

Where $C(\theta)$ is the parametrization of a circular contour using the variable theta. Fitting our function to the parametrization and evaluating: \begin{aligned} \oint_{C}{f(z)\ \mathrm{d}z} &= \int_0^{\pi}{\frac{ie^{i \theta}}{R^2 e^{2it} + 1}}\ \mathrm{d}\theta \\&= -\frac{2 \tan^{-1}(R)}{R} \end{aligned}

We know that the ArcTangent is bounded by [0, 1) for all $0 \leq \theta \leq \pi$ so we can say $-\frac{2 \tan^{-1}(R)}{R} \leq -\frac{2M}{R}$. Next apply the limit for $R \to 0$

$$\lim_{R \to \infty}{ -\frac{2M}{R}} = 0$$

Thus, we have shown that as $R \to \infty$, the integral around the contour "vanishes".

This may go nowhere but perhaps you can even apply Cauchy-Goursat theorem to show that the integral about the contour is 0 so long as $i$ is not in the region enclosed by the contour. i.e. $R < 1$. Otherwise, when $R > 1$ use deformation of contours. Something to think about.

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