I'm trying to evaluate the real integral $$\int_{ - \infty}^{\infty} \frac{dx}{1+x^2}$$

Denote $\Gamma_{1}=\left[-R,R\right]\ \Gamma_{2}=Re^{it}$, for $t\in\left[0,\pi\right]$, and let $\gamma$ be a small circle around $i$ so $\gamma$ is inside the area bounded by $\Gamma_{1}\cup\Gamma_{2}$. By Cauchy's theorem: $$ \int_{\Gamma_{1}}f\left(z\right)dz+\int_{\Gamma_{2}}f\left(z\right)dz=\int_{\gamma}f\left(z\right)dz $$ And calculating $\int_{\gamma}f\left(z\right)dz$ gives us $\pi$ (operating Cauchy's formula on the function $\frac{1}{z+i}$). so we got $$\int_{\Gamma_{1}}f\left(z\right)dz+\int_{\Gamma_{2}}f\left(z\right)dz=\pi$$ now I need to show that $$\lim_{R\rightarrow\infty}\int_{\Gamma_{2}}f\left(z\right)dz=0$$ and I'm stuck.

  • 1
    $\begingroup$ Search and read about Jordan's lemma. $\endgroup$
    – Taozi
    May 5, 2016 at 7:42

1 Answer 1


You can apply Estimation lemma. Since $$ \left|\int_{\Gamma_2} \frac{1}{1+z^2}dz\right| \le \frac{\pi R}{R^2 -1} $$ for large $R$, $$ \lim_{R\to\infty}\left|\int_{\Gamma_2} \frac{1}{1+z^2}dz\right|=0. $$ Then you can get what you want.

  • $\begingroup$ As simple as it is beautiful! Thank you! $\endgroup$
    – Uria Mor
    May 5, 2016 at 7:48

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