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Thread: [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 $\displaystyle \int _0^{\infty} \frac{x^2dx}{(x^2+1)(x^2+4)}=\frac{\pi}{6}$.

    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...

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

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



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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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...

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

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



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

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

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

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

    Kind regards

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


    Also, is $\displaystyle \lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz=0$ equal to $\displaystyle \lim_{R \rightarrow \infty} \int_{-R}^R f(z)\cdot dz$?
    I don't think so, as there is 2 singularities as you pointed out. I think I'm misunderstanding something.
    So I should calculate $\displaystyle \int _ {\gamma _1} f(z) dz$ where $\displaystyle \gamma_1$ is a closed path around $\displaystyle i$? Then I sum this result to $\displaystyle \int _ {\gamma _2} f(z) dz$ where $\displaystyle \gamma _2$ is a closed path around $\displaystyle 2i$. 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...

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

    ... where...

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

    In our example is...

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

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



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

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

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

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

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

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

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

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

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

    ... so that...

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


    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Oct 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...

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

    ... where...

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

    In our example is...

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

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



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

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

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

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

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

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

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

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

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

    ... so that...

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


    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    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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