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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?

  • 1
    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.2012-05-10
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    $$\left|\frac{1}{1+z^2}\right| \le \frac{1}{|z|^2-1}$$ if $|z| > 1$.2012-05-10

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