I'm trying to compute the following integral using the residue theorem: $$\int_{-\infty}^\infty \frac{dx}{x^2-2x+4}$$

I know that this integral is equal to the integral over the upper half plane since the integral over the semi-circle in the upper half plane goes to zero as $R \to \infty$. I'm happy with computing this integral and have found the result to be $\pi/\sqrt{3}$ (which I'm fairly confident is correct).

However, I'm having trouble showing that the integral over the semi-circle in the upper half plane does go to zero. I know that I can use the $Ml$ inequality to show this.

Let the radius of the semi-circle be $R$. So I know that $l$ is $2\pi R$. So then I need to find an upper bound on the modulus of the integrand. This is where I'm stuck. How do I do this?

  • 1
    $\begingroup$ Parametrise $x = Re^{i\theta}$ and use the triangle inequality to make some estimations. $\endgroup$ – ÍgjøgnumMeg Dec 23 '16 at 20:53
  • $\begingroup$ Because the denominator is at least two degrees higher than the numerator, you know the semi-circular contour drops to zero $\endgroup$ – Kaynex Dec 23 '16 at 21:03

You maximal value is on the boundary of your semidisk. Take R o infinity and you are ready. $$\frac{1}{2\pi }\left|\int_{\gamma}\frac{1}{z^2-2z+4}\right|\le \frac{2\pi R}{2\pi}\max_{|z|=R}\left|\frac{1}{z^2-2z+4}\right|\le\frac{R}{R^2-2R+4}\le 1/R \to 0 $$ as $R\to \infty$


Complex plane

In the complex plane we use the function $$ f(z) = \frac{1}{z^{2}-2z+4} $$

complex plane

Find poles

Where does $z^{2}-2z+4=0$? When $$ z = 1 \pm i\sqrt{3} $$ These poles are represented as $\color{red}{\times}$ in the figure below.



Jordan lemma

$$ \oint f(z)\, dz = \lim_{R\to\infty} \left( \int_{\Gamma_{R}} f(z)\, dz + \int_{\Omega_{R}} f(z)\, dz \right) $$ Therefore, $$ \int_{-\infty}^{\infty} f(x)\, dx = \oint f(z)\, dz + \lim_{R\to\infty} \left( \int_{\Gamma_{R}} f(z)\, dz \right) $$

Because $$ \lim_{z\to\infty} |z\, f(z)| = \lim_{z\to\infty} \Bigg|\frac{z}{z^{2}-2z+4}\Bigg| = 0 $$ $$ \lim_{R\to\infty} \left( \int_{\Gamma_{R}} f(z)\, dz \right) = 0 $$

Residue theorem

$$ \oint_{\Gamma_{R}} f(z)\, dz = 2\pi i \sum_{k} \text{Res }f(z_{k}) $$ where the points $z_{k}$ are the poles enclosed by the contour. In this instance, there is a single pole at $$z_{1} = 1+i\sqrt{3}$$ The Laurent expansion about this point is $$ f\left(z-z_{1}\right) = -\frac{i}{2 \sqrt{3}} \frac{1}{\zeta -1 - \sqrt{3}} + \frac{1}{12} + \frac{i}{24 \sqrt{3}} \left(\zeta -1-i \sqrt{3}\right) + \mathcal{O}\left(\zeta -1-i \sqrt{3} \right)^{2} $$ $$ \oint_{\Gamma_{R}} f(z)\, dz = 2\pi i \left( -\frac{i}{2 \sqrt{3}} \right) = \frac{\pi}{\sqrt{3}} $$


$$ \int_{\infty}^{\infty} \frac{dx}{ x^{2}-2x+4 } = \frac{\pi}{\sqrt{3}} $$

  • $\begingroup$ We don't need the Jordan's lemma here ? $\endgroup$ – Zaid Alyafeai Apr 13 '17 at 23:03

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