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Math Help - Fourier Sine Integral (Complex Analysis)

  1. #1
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    Fourier Sine Integral (Complex Analysis)

    This question is similar to another one I've posted

    Show:
    \int_{-\infty}^{\infty} \frac{\sin{x}}{x^2+4x+5} = \frac{-\pi\sin{2}}{e}

    so I let f(z) = \frac{\sin{z}}{z^2+4z+5}
    And I used a closed semicircle which contains  z = -2+i of radius R.

    Let C_{R} = \{z=Re^{i\theta}: 0\leq\theta\leq\pi\, R<\infty\}
    C_{+} = \{z=r: -R<r<R, R<\infty\}

    I let: I = \int_{C_{+}} f(z)dz and I_{R} = \int_{C_R} f(z)dz

    By Jordan's Lemma, I_{R} = 0 as R\to\infty

    By the Cauchy Residue Theorem
    <br />
I + I_{R} = I = 2\pi i {Res}_{z=-2+i} f(z)

    Now I use the following theorem to evaluate the theorem
    If p(z) is entire and q(z) is analytic everywhere except z_{0} but q'(z) is entire.

    The {Res}_{z=z_0} \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}

    However, I can't get the following to evaluate correctly

    I = 2\pi i \frac{\sin{(-2+i)}}{2(-2+i) + 4(-2+i)}

    I am having troubles with  \sin{(-2+i)}. None of the identities in my text book seem to help.
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  2. #2
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    Quote Originally Posted by Haven View Post
    This question is similar to another one I've posted

    Show:
    \int_{-\infty}^{\infty} \frac{\sin{x}}{x^2+4x+5} = \frac{-\pi\sin{2}}{e}

    so I let f(z) = \frac{\sin{z}}{z^2+4z+5}
    And I used a closed semicircle which contains  z = -2+i of radius R.

    Let C_{R} = \{z=Re^{i\theta}: 0\leq\theta\leq\pi\, R<\infty\}
    C_{+} = \{z=r: -R<r<R, R<\infty\}

    I let: I = \int_{C_{+}} f(z)dz and I_{R} = \int_{C_R} f(z)dz

    By Jordan's Lemma, I_{R} = 0 as R\to\infty

    By the Cauchy Residue Theorem
    <br />
I + I_{R} = I = 2\pi i {Res}_{z=-2+i} f(z)

    Now I use the following theorem to evaluate the theorem
    If p(z) is entire and q(z) is analytic everywhere except z_{0} but q'(z) is entire.

    The {Res}_{z=z_0} \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}

    However, I can't get the following to evaluate correctly

    I = 2\pi i \frac{\sin{(-2+i)}}{2(-2+i) + 4(-2+i)}

    I am having troubles with  \sin{(-2+i)}. None of the identities in my text book seem to help.

    First remember that

    e^{iz}=\cos(z)+i\sin(z)

    So

    \oint \frac{\sin(z)}{z^2+4z+5}dz= Im\left\{\oint\frac{e^{iz}}{(z^2+4z+5)}dz \right\}

    let g(z)=\frac{e^{iz}}{(z^2+4z+5)}

    Then

    Res(g,z=-2+i)=\frac{e^{-1-2i}}{2i}=-i\frac{\cos(2)}{2e}-\frac{\sin(2)}{2e}

    \oint \frac{e^{iz}}{z^2+4z+5}dz=2\pi i \left(-i\frac{\cos(2)}{2e}-\frac{\sin(2)}{2e} \right)=\frac{\pi\cos(2)}{e}-i\frac{\pi\sin(2)}{e}

    Now we know that the integral above is the imaginary part of this so

    \int_{-\infty}^{\infty} \frac{\sin(z)}{z^2+4z+5}dz=-\frac{\pi\sin(2)}{e}
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