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Math Help - [SOLVED] Solving an integral via complex analysis

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Solving an integral via complex analysis

    I must prove that \int _0^{\infty} \frac{x^2dx}{(x^2+1)(x^2+4)}=\frac{\pi}{6}.

    My attempt : None, I don't know where to start. I'm really lost, I don't have any textbook but my classnotes.
    I'd like a push in the right direction rather than a full answer, but everything's welcome.
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  2. #2
    MHF Contributor chisigma's Avatar
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    A possible solution is to integrate, using the residue theorem, the function...

    f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}

    ... along the grey path of the figure...



    ... and let R tend to infinity. By Jordan's lemma is...

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

    ... and the singularities of f(*) inside the path are z=i and z=2i. Since f(x) is an even function respect to x it will be...

    \int_{0}^{\infty} f(x)\cdot dx = \frac{1}{2} \int_{-\infty}^{+\infty} f(x)\cdot dx

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by chisigma View Post
    A possible solution is to integrate, using the residue theorem, the function...

    f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}

    ... along the grey path of the figure...



    ... and let R tend to infinity. By Jordan's lemma is...

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

    ... and the singularities of f(*) inside the path are z=i and z=2i. Since f(x) is an even function respect to x it will be...

    \int_{0}^{\infty} f(x)\cdot dx = \frac{1}{2} \int_{-\infty}^{+\infty} f(x)\cdot dx

    Kind regards

    \chi \sigma
    Moltissimo grazie, ma... why do you write f(*) and not f(z)?


    Also, is \lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz=0 equal to \lim_{R \rightarrow \infty} \int_{-R}^R f(z)\cdot dz?
    I don't think so, as there is 2 singularities as you pointed out. I think I'm misunderstanding something.
    So I should calculate \int _ {\gamma _1} f(z) dz where \gamma_1 is a closed path around i? Then I sum this result to \int _ {\gamma _2} f(z) dz where \gamma _2 is a closed path around 2i. Unless I'm not getting it.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The residue theorem states that, if C is a closed path and a function f(*) has a finite number n of simple poles inside C, is...

    \int_{C} f(z) = 2\pi i \sum_{i=1}^{n} r_{i} (1)

    ... where...

    r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i}) f(z) (2)

    In our example is...

    f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)} (3)

    ... C is the gray path in figure...



    ... for the Jordan's lermma is...

    \lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz= 0 (4)

    ... so that, combining (1),(2),(3) and (4), is...

    \int_{-\infty}^{\infty}\frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)}= 2\pi i\cdot \sum_{i=1}^{n} r_{i} (5)

    In our case f(*) has only two poles inside C : z_{1}=i, z_{2}=2i, so that is...

    r_{1}= \lim_{z \rightarrow i} (z-i)\cdot f(z)= \frac{i}{6}

    r_{2}= \lim_{z \rightarrow 2i} (z-2i)\cdot f(z)= -\frac{i}{3} (6)

    By (1) and (5) we have finally...

    \int_{-\infty}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{3} (7)

    ... so that...

    \int_{0}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{6} (8)


    Kind regards

    \chi \sigma
    Last edited by chisigma; October 18th 2009 at 10:55 PM.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by chisigma View Post
    The residue theorem states that, if C is a closed path and a function f(*) has a finite number n of simple poles inside C, is...

    \int_{C} f(z) = 2\pi i \sum_{i=1}^{n} r_{i} (1)

    ... where...

    r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i}) f(z) (2)

    In our example is...

    f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)} (3)

    ... C is the gray path in figure...



    ... for the Jordan's lermma is...

    \lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz= 0 (4)

    ... so that, combining (1),(2),(3) and (4), is...

    \int_{-\infty}^{\infty}\frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)}= 2\pi i\cdot \sum_{i=1}^{n} r_{i} (5)

    In our case f(*) has only two poles inside C : z_{1}=i, z_{2}=2i, so that is...

    r_{1}= \lim_{z \rightarrow i} (z-i)\cdot f(z)= \frac{i}{6}

    r_{2}= \lim_{z \rightarrow 2i} (z-2i)\cdot f(z)= -\frac{i}{3} (6)

    By (1) and (5) we have finally...

    \int_{-\infty}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{3} (7)

    ... so that...

    \int_{0}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{6} (8)


    Kind regards

    \chi \sigma
    Moltissimo grazie di nuovo!
    Very well explained.
    If you don't mind me asking a last question : Are these type of integrals common in physics?
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