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June 8th, 2019, 08:27 AM   #1
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how to solve the integral with residues?
If I can solve it like that
Attached Images 2ubkBV1Pllo.jpg (93.9 KB, 20 views) November 1st, 2019, 03:41 PM #2 Newbie   Joined: Jun 2016 From: UK Posts: 8 Thanks: 8 Math Focus: Anything but stats :) In order to evaluate $I \displaystyle = \int_{-\infty}^{\infty} \dfrac{dx}{x^2+4}$ using residues, we first need to consider the complex contour integration given by $I_C = \displaystyle \int_C \dfrac{dz}{z^2 + 4}$ where $C$ is a CLOSED contour representing a semicircle lying in the upper-half of the complex plane, centred at the origin and of radius $R>0$. This contour integral can be decomposed into two simpler ones: (a) A contour integral along the real line from $-R$ to $R$. This is represented as $\displaystyle \int_{-R}^R \dfrac{dz}{z^2 + 4}$. Note that on the real line, we have $z=x$ therefore this is literally just the same as $\displaystyle \int_{-R}^R \dfrac{dx}{x^2 + 4}$. and (b) A contour integral along the semicircle in the upper half plane. This is represented by $\displaystyle \int_{\Gamma} \dfrac{dz}{z^2 + 4}$ where $\Gamma : z=Re^{i\theta}$ where $\theta \in (0,\pi)$. Therefore, we have that $\displaystyle I_C = \int_{-R}^R \dfrac{dx}{x^2 + 4} + \int_\Gamma \dfrac{dz}{z^2 + 4}$. Now consider both sides in the limit $R \to \infty$. We get that $\boxed{ \displaystyle \lim_{R \to \infty} I_C = \int_{-\infty}^{\infty} \dfrac{dx}{x^2 + 4} + \lim_{R \to \infty}\int_\Gamma \dfrac{dz}{z^2 + 4} }$ and as you can see, the integral we want is literally just sitting there. Luckily, we can evaluate the other two terms with some introductory theory. Firstly, note that in the second integral, we have $z = Re^{i \theta}$ where $\theta \in (0,\pi)$. Also note that $|z^2 + 4| \geq |z^2| + 4 = R^2 + 4$ therefore $\dfrac{1}{|z^2 + 4|} \leq \dfrac{1}{R^2 + 4}$. This is important because we can use something known as Estimation Lemma (or ML inequality) to get an upper bound on our integral. Therefore, notice that $\displaystyle \left| \int_\Gamma \dfrac{dz}{z^2 + 4} \right| \leq \dfrac{1}{R^2 + 4}L$ where $L$ is the length of our contour. In our case, since the contour is a semicircle, $L = \pi R$. Therefore $\displaystyle \left| \int_\Gamma \dfrac{dz}{z^2 + 4} \right| \leq \dfrac{\pi R}{R^2 + 4}$ and as we take the limit $R \to \infty$, the upper bound goes to zero therefore this entire integral vanishes, i.e. $\displaystyle \lim_{R \to \infty} \int_{\Gamma} \dfrac{dz}{z^2+4} = 0$. Secondly, to work out $\displaystyle \lim_{R \to \infty} I_C$ we note that $I_C$ is a closed contour and so we use what's known as Residue Theorem whereby the value of this term is precisely the $2\pi i$ multiplied by the sum of all residues of $\dfrac{1}{z^2 + 4}$. There is only one residue for this function in our contour, and this is precisely $z_0 = 2i$. Therefore, we have that $\displaystyle \lim_{R \to \infty} I_C = 2\pi i \mathrm{Res}_{z = 2i} \dfrac{1}{z^2 + 4} = 2\pi i \lim_{z \to 2i} (z-2i)\dfrac{1}{z^2 + 4} = 2\pi i \lim_{z \to 2i}\dfrac{1}{z+2i} = 2\pi i \dfrac{1}{4i} = \dfrac{\pi}{2}$ Therefore, our boxed expression reduces (in the limit) down to $\dfrac{\pi}{2} = \displaystyle \int_{-\infty}^{\infty} \dfrac{dx}{x^2 + 4}$ And so coming back to your problem, your integral is precisely half this, therefore $\dfrac{\pi}{8}$. November 4th, 2019, 11:10 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,114 Thanks: 2329 $\displaystyle \frac12\left(\frac{\pi}{2}\right) = \frac{\pi}{4} \text{, not }\frac{\pi}{8}$. Tags integral Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jhenrique Calculus 5 June 30th, 2015 04:45 PM gen_shao Calculus 2 July 31st, 2013 10:54 PM maximus101 Calculus 0 March 4th, 2011 02:31 AM xsw001 Real Analysis 1 October 29th, 2010 08:27 PM maximus101 Algebra 0 December 31st, 1969 04:00 PM

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