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Math Help - Hard complex analysis question help Please!

  1. #1
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    Hard complex analysis question help Please!

    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
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  2. #2
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    Quote Originally Posted by clairesina View Post
    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
    Perhaps start by noting that \sin^2 x = \frac{1}{2} \left( 1 - \cos (2x) \right) so maybe integrate \frac{1 - e^{2iz}}{z^2} = \frac{1}{z^2} - \frac{e^{2iz}}{z^2} around an appropriate contour.
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  3. #3
    MHF Contributor chisigma's Avatar
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    I remember that the expression for cos(*) in term of exponential is...

    \cos x = \frac{e^{ix} + e^{-ix}}{2}= \cosh ix

    ... so that ...

    \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...

    \frac{1 - \cosh 2iz}{2\cdot z^{2}}

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 15th 2009 at 11:46 AM. Reason: corrections
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  4. #4
    MHF Contributor chisigma's Avatar
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    Today we will try to resolve the indefinite integral...

    \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

    \int_{C} f(z)\cdot dz= 2 \pi i\cdot \sum_{k} R_{k} (2)

    where R_{k} is the residue of f(*) in one singularity a_{k} of order n internal to C

    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...

    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...

    \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 |f(z)|\le \frac{M}{R^{k}} for z=R\cdot e^{i \theta}, then is...

    \lim_{R\rightarrow \infty} \int_{ABC} f(z)\cdot dz=0 (6)

    ... and that for the f(*) given in (4) is verified for M= \frac{1}{2} and k=2... very well!... In that case, taking into account (5) and (6), we can write...

    \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

     = 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...

    \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

    \chi \sigma
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  5. #5
    MHF Contributor chisigma's Avatar
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    In my previous post the attempt to compute the definite integral…

    \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…

    \int_{C} f(z)\cdot dz= 2 \pi i\cdot \sum_{k} R_{k} (2)

    … where R_{k} is the residue of f(*) in one singularity a_{k} of order n internal to C…

    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...

    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...

    \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 |f(z)|\le \frac{M}{R^{k}} for z=R\cdot e^{i \theta}, then is...

    \lim_{R\rightarrow \infty} \int_{ABC} f(z)\cdot dz=0 (6)

    ... and that for the f(*) given in (4) is verified for M= \frac{1}{2} and k=2... very well!... In that case, taking into account (5) and (6), we can write...

    \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

     = 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...

    \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

    \chi \sigma
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  6. #6
    MHF Contributor chisigma's Avatar
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    Now we perform the final step, that is the computation of...

    \int_{- \infty}^{+\infty} \frac{\sin^{2} x}{x^{2}}\cdot dx = \lim_{r\rightarrow 0} \int_{AB} f(z) \cdot dz (1)

    ... where...

    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 z=r\cdot e^{i \theta}, so that is dz= i \cdot r \cdot e^{i \theta} and the integral becomes...

    \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...

    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...

    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...

    \lim_{r \rightarrow 0} g(\theta) =1 (6)

    ... so that...

    \int_{- \infty}^{+ \infty} \frac{\sin^{2} x}{x^{2}} \cdot dx = \int_{0}^{\pi} d\theta = \pi (7)

    Kind regards

    \chi \sigma
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