My Math Forum  

Go Back   My Math Forum > College Math Forum > Complex Analysis

Complex Analysis Complex Analysis Math Forum


Reply
 
LinkBack Thread Tools Display Modes
June 8th, 2019, 08:27 AM   #1
Newbie
 
Joined: Jun 2019
From: Ukraine

Posts: 6
Thanks: 0

Question Integral

how to solve the integral with residues?
If I can solve it like that
Attached Images
File Type: jpg 2ubkBV1Pllo.jpg (93.9 KB, 20 views)
2pacman is offline  
 
November 1st, 2019, 03:41 PM   #2
Newbie
 
Joined: Jun 2016
From: UK

Posts: 8
Thanks: 8

Math Focus: Anything but stats :)
Smile

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 $\dfrac{\pi}{8}$.
RDKGames is offline  
November 4th, 2019, 11:10 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 21,114
Thanks: 2329

$\displaystyle \frac12\left(\frac{\pi}{2}\right) = \frac{\pi}{4} \text{, not }\frac{\pi}{8}$.
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Complex Analysis

Tags
integral



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Double integral, repeated integral and the FTC Jhenrique Calculus 5 June 30th, 2015 04:45 PM
a new discovery in integral calculus??for integral pros only gen_shao Calculus 2 July 31st, 2013 10:54 PM
integral of double integral in a region E maximus101 Calculus 0 March 4th, 2011 02:31 AM
Prove Lower Integral <= 0 <= Upper Integral xsw001 Real Analysis 1 October 29th, 2010 08:27 PM
integral of double integral in a region E maximus101 Algebra 0 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.