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Math Help - Solving integral using residue

  1. #1
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    Solving integral using residue

    Need help to complete the following integral using the residue theorem:


    \displaystyle\int_{0}^{\infty}\frac{cos x}{x^{4}+4}dx= \frac{1}{2}\int_{-\infty}^{\infty}\frac{cos x}{x^{4}+4}dx=\frac{1}{2}Re\int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+4}dx\displayst  yle\implies

    \displaystyle\implies f(z)=\frac{e^{iz}}{z^{4}+4}=\frac{e^{iz}}{(z^{2}+2  i)(z^{2}-2i)}

    \displaystyle{z_{1}=(1+i)
    \displaystyle{z_{2}=(-1+i)
    \displaystyle{z_{3}=(-1-i)
    \displaystyle{z_{4}=(-1-i)

    The residue at z_1 is:

    \displaystyle\frac{1}{2} \mathrm{Res}_{z \to z_1} f(z)\lim_{z \to z_1}=\frac{e^{iz}}{4z^{3}}= \frac{1}{4} \frac{e^{iz}}{z^{3}}|_{z=z_1}=(-1-i)\frac{e^{(i-1)}}{32} (1)

    The residue at z_2 is:

    \displaystyle\frac{1}{2} \mathrm{Res}_{z \to z_2} f(z)\lim_{z \to  z_2}=\frac{e^{iz}}{4z^{3}}= \frac{1}{4}  \frac{e^{iz}}{z^{3}}|_{z=z_2}=(1-i)\frac{e^{(-i-1)}}{32} (2)

    \displaystyle I=2\pi i (((-1-i)\frac{e^{(i-1)}}{32})+((1-i)\frac{e^{(-i-1)}}{32}))=

    \displaystyle=\pi i (((-1-i)\frac{e^{(i-1)}}{16})+((1-i)\frac{e^{(-i-1)}}{16}))=... (3)

    How do I simplify (3)? DO I take the real part of the expression
    (3)? In that case how do I do that?

    I appriciate any guidance.

    Thank you

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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by 4Math View Post
    \displaystyle{z_{4}=(-1-i)
    It should be z_4=1-i.

    The residue at z_1 is:

    \displaystyle\frac{1}{2} \mathrm{Res}_{z \to z_1} f(z)\lim_{z \to z_1}=\frac{e^{iz}}{4z^{3}}= \frac{1}{4} \frac{e^{iz}}{z^{3}}|_{z=z_1}=(-1-i)\frac{e^{(i-1)}}{32}
    I don't know what are you doing here. We have:

    \textrm{Res}[f,z_1]=\displaystyle\lim_{z \to z_1}{\dfrac{e^{iz}}{(z-z_2)(z-z_3)(z-z_4)}}=\\<br />
\dfrac{e^{iz_1}}{(z_1-z_2)(z_1-z_3)(z_1-z_4)}=\ldots

    Regards.

    Fernando Revilla
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    It should be z_4=1-i
    Thank you for the correction.


    I don't know what are you doing here.
    I initially used the following:
    \displaystyle\mathrm{Res}_{z \to a} \frac{f_1(z)}{f_2(z)}=\frac{f_1(a)}{f{'}_2(a)} there f_2 has a single solution z=a. But I realized that I can't apply this since there are two possible solutions in the north region z_1 and z_2.

    We have:

    \textrm{Res}[f,z_1]=\displaystyle\lim_{z \to z_1}{\dfrac{e^{iz}}{(z-z_2)(z-z_3)(z-z_4)}}=\\<br />
\dfrac{e^{iz_1}}{(z_1-z_2)(z_1-z_3)(z_1-z_4)}=\ldots
    Can you please clarify and generalize this rule for this kind of problem ( which theorem etc...)

    \textrm{Res}[f,z_1]=\displaystyle\lim_{z \to z_1}{\dfrac{e^{iz}}{(z-z_2)(z-z_3)(z-z_4)}}=\\\dfrac{e^{iz_1}}{(z_1-z_2)(z_1-z_3)(z_1-z_4)}=

    \displaystyle=(-1-i)\\\dfrac{e^{(i-1)}}{8(-1+i)(-1-i)}=(-1-i)\\\dfrac{e^{(i-1)}}{16} (4)
    \textrm{Res}[f,z_2]=\displaystyle\lim_{z \to z_2}=\\\dfrac{e^{iz_2}}{(z_2-z_1)(z_2-z_3)(z_2-z_4)}=

    =(1-i)\\\dfrac{e^{(-1-i)}}{16}(5)

    \displaystyle I=2\pi i \frac{1}{2}(((-1-i)\\\dfrac{e^{(i-1)}}{16})+((1-i)\\\dfrac{e^{(-1-i)}}{16}))=

    \displaystyle=\pi i \frac{1}{2}(\\\dfrac{(-1-i)e^{i}+((1-i)e^{-i})}{8e}))=

    \displaystyle=\pi \frac{1}{2}(\\\dfrac{(-ie^{i}+e^{i}+ie^{-i}+e^{-i})}{8e})=

    \displaystyle=\frac{1}{2}(-ie^{i}+e^{i}+ie^{-i}+e^{-i})(\frac{\pi}{8e}) (6)

    \displaystyle\frac{1}{2}(e^{i}+e^{-i})= cos 1

    \displaystyle\frac{1}{2}(ie^{-i}-ie^{i})=\frac{i}{2}(e^{-i}-e^{i})=\frac{-1}{2i}(e^{-i}-e^{i})=\frac{1}{2i}(e^{i}-e^{-i})= sin 1

    \displaystyle\implies\int_{0}^{\infty}\frac{cos x}{x^{4}+4}dx=\frac{\pi}{8e}(cos 1 + sin1)

    Is the answer correct?

    Thank you

    Kind regards
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by 4Math View Post
    Can you please clarify and generalize this rule for this kind of problem ( which theorem etc...)
    We only use the decomposition of z^4+4:

    \textrm{Res}[f,z_1]=\displaystyle\lim_{z \to z_1}{\dfrac{e^{iz}\cancel{(z-z_1)}}{\cancel{(z-z_1})(z-z_2)(z-z_3)(z-z_4)}}=

    \displaystyle\lim_{z \to z_1}{\dfrac{e^{iz}}{(z-z_2)(z-z_3)(z-z_4)}}=<br />
\dfrac{e^{iz_1}}{(z_1-z_2)(z_1-z_3)(z_1-z_4)}=\ldots


    Is the answer correct?
    Completely correct.

    Regards.

    Fernando Revilla
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