Hi,
I totally have no idea how to do this question which involves residue...
Intergral from negative infinity to positive infinity of
((Sinx)^2 / X^2) dx
Thank you so much!
Claire
I remember that the expression for cos(*) in term of exponential is...
$\displaystyle \cos x = \frac{e^{ix} + e^{-ix}}{2}= \cosh ix$
... so that ...
$\displaystyle \frac {\sin^{2} x}{x^{2}}= \frac{1-\cos 2x}{2\cdot x^{2}} = \frac{1 - \cosh 2ix}{2\cdot x^{2}} $
... and the function to integrate may be the following...
$\displaystyle \frac{1 - \cosh 2iz}{2\cdot z^{2}}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Today we will try to resolve the indefinite integral...
$\displaystyle \int_{-\infty}^{+\infty} \frac{\sin^{2} x}{x^{2}}\cdot dx$ (1)
... using the residue theorem. This theorem gives the integral of a complex variable function f(*) along a closed path C as…
$\displaystyle \int_{C} f(z)\cdot dz= 2 \pi i\cdot \sum_{k} R_{k}$ (2)
… where $\displaystyle R_{k}$ is the residue of f(*) in one singularity $\displaystyle a_{k}$ of order n internal to C…
$\displaystyle R_{k}= \lim_{z\rightarrow a_{k}} \frac {1}{(n-1)!} \cdot \frac{d^{n-1}}{dz^{n-1}} \{(z-a_{k})^{n}\cdot f(z)\}$ (3)
The f(*) we choose is...
$\displaystyle f(z)=\frac{1-\cosh 2iz}{2\cdot z^{2}}$ (4)
... and the path C is indicated in figure as ABCDEF...
Since there are no singulatities internal to the path is...
$\displaystyle \int_{C} f(z)\cdot dz = \int_{ABC} f(z)\cdot dz + \int_{CD} f(z)\cdot dz + \int_{DF} f(z)\cdot dz + \int_{FA} f(z)\cdot dz =0$ (5)
First we will consider the integral along ABC, to which we can apply the so called 'Jordan's Lemma' that says that if there are two constants M and k>1 so that is $\displaystyle |f(z)|\le \frac{M}{R^{k}}$ for $\displaystyle z=R\cdot e^{i \theta}$, then is...
$\displaystyle \lim_{R\rightarrow \infty} \int_{ABC} f(z)\cdot dz=0$ (6)
... and that for the f(*) given in (4) is verified for $\displaystyle M= \frac{1}{2}$ and $\displaystyle k=2$... very well!... In that case, taking into account (5) and (6), we can write...
$\displaystyle \lim_{R\rightarrow \infty} \int_{c} f(z)\cdot dz= \int_{r}^{\infty} \frac{1-\cos 2x}{x^{2}}\cdot dx + \int_{DF} f(z)\cdot dz $
$\displaystyle = 2\cdot \int_{r}^{\infty}\frac{\sin^{2} x}{x^{2}}\cdot dx + \int_{DF} f(z)\cdot dz = 0$ (7)
... or in similar way...
$\displaystyle \int_{-\infty}^{+ \infty}\frac{\sin^{2} x}{x^{2}}\cdot dx = -\lim_{r\rightarrow 0} \int_{DF} f(z)\cdot dz$ (8)
... and this limit will be computed in next post...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
In my previous post the attempt to compute the definite integral…
$\displaystyle \int_{-\infty}^{+\infty} \frac{\sin^{2} x}{x^{2}}\cdot dx$ (1)
... has failed because the function I proposed didn’t satisfy the ‘Jordan’s Lemma’. So I will correct the post and please the moderators, if possible, to delete my previous post in order to avoid misunderstanding…
The so called ‘Residue Theorem’ gives the integral of a complex variable function f(*) along a closed path C as…
$\displaystyle \int_{C} f(z)\cdot dz= 2 \pi i\cdot \sum_{k} R_{k}$ (2)
… where $\displaystyle R_{k}$ is the residue of f(*) in one singularity $\displaystyle a_{k}$ of order n internal to C…
$\displaystyle R_{k}= \lim_{z\rightarrow a_{k}} \frac {1}{(n-1)!} \cdot \frac{d^{n-1}}{dz^{n-1}} \{(z-a_{k})^{n}\cdot f(z)\}$ (3)
The f(*) we choose is the function proposed by Fantastic...
$\displaystyle f(z)=\frac{1-e^{2iz}}{2\cdot z^{2}}$ (4)
... and the path C is indicated in figure as ABCDEF...
Since there are no singulatities internal to the path is...
$\displaystyle \int_{C} f(z)\cdot dz = \int_{ABC} f(z)\cdot dz + \int_{CD} f(z)\cdot dz + \int_{DF} f(z)\cdot dz + \int_{FA} f(z)\cdot dz =0$ (5)
First we will consider the integral along ABC, to which we can apply the so called 'Jordan's Lemma' that says that if there are two constants M and k>1 so that is $\displaystyle |f(z)|\le \frac{M}{R^{k}}$ for $\displaystyle z=R\cdot e^{i \theta}$, then is...
$\displaystyle \lim_{R\rightarrow \infty} \int_{ABC} f(z)\cdot dz=0$ (6)
... and that for the f(*) given in (4) is verified for $\displaystyle M= \frac{1}{2}$ and $\displaystyle k=2$... very well!... In that case, taking into account (5) and (6), we can write...
$\displaystyle \lim_{R\rightarrow \infty} \int_{c} f(z)\cdot dz= \int_{-\infty}^{-r} \frac{1- e^{2ix}}{2\cdot x^{2}}\cdot dx + \int_{DF} f(z)\cdot dz + \int_{r}^{+ \infty} \frac{1- e^{2ix}}{2\cdot x^{2}}\cdot dx $
$\displaystyle = 2\cdot \int_{r}^{\infty}\frac{\sin^{2} x}{x^{2}}\cdot dx + \int_{DF} f(z)\cdot dz = 0$ (7)
... or in similar way...
$\displaystyle \int_{-\infty}^{+ \infty}\frac{\sin^{2} x}{x^{2}}\cdot dx = -\lim_{r\rightarrow 0} \int_{DF} f(z)\cdot dz$ (8)
... and this limit will be computed in next post...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Now we perform the final step, that is the computation of...
$\displaystyle \int_{- \infty}^{+\infty} \frac{\sin^{2} x}{x^{2}}\cdot dx = \lim_{r\rightarrow 0} \int_{AB} f(z) \cdot dz$ (1)
... where...
$\displaystyle f(z)= \frac{1-e^{2iz}}{2\cdot z^{2}}$ (2)
... and the path AB is represented in the figure...
For the computation of (1) we set $\displaystyle z=r\cdot e^{i \theta}$, so that is $\displaystyle dz= i \cdot r \cdot e^{i \theta}$ and the integral becomes...
$\displaystyle \int_{0}^{\pi} \frac {1-e^{2\cdot i \cdot r \cdot e^{i \theta}}}{2 \cdot r \cdot e^{i \theta}} \cdot d \theta = \int_{0}^{\pi} g(\theta) \cdot d \theta$ (3)
Using the Taylor expansion of exponential...
$\displaystyle e^{2\cdot i \cdot r \cdot e^{i \theta}} = \sum_{n=0}^{\infty} \frac{(2 \cdot i \cdot r \cdot e^{i \theta})^{n}}{n!}$ (4)
... we obtain the function g(*) that must be integrated in the form...
$\displaystyle g(\theta) = 1 + i \cdot r \cdot e^{i \theta} + \frac{2}{3} \cdot i \cdot r^{2} \cdot e^{2 i \theta} - \dots$ (5)
... and what really matters is that...
$\displaystyle \lim_{r \rightarrow 0} g(\theta) =1$ (6)
... so that...
$\displaystyle \int_{- \infty}^{+ \infty} \frac{\sin^{2} x}{x^{2}} \cdot dx = \int_{0}^{\pi} d\theta = \pi$ (7)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$