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Math Help - Complex Analysis - Integration

  1. #1
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    Complex Analysis - Integration

    Use a theorem to show that:

    \int_{-\infty}^{\infty}3sin2t/t(t^2+9)dt = (\pi i/3)(1-1/e^6)
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  2. #2
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    How can the integral that you gave, which is real valued possibly have i in it? Check your problem again.
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  3. #3
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    I was moving threads and found this one.
    Let me try to find this integral.

    Here we need to use two theorems.

    Michael Jordan's Lemma: Let f(z) be a meromorphic function in the upper half-plane with finitely many poles: b_1,...,b_n in upper half-plane. Given that f(z)\to 0 \mbox{ as }|z| \to \infty in the upper-half plane. Then for each positive number a we have:
    \int_{-\infty}^{\infty} e^{iax} f(x) dx = 2\pi i \sum_{k=1}^n \mbox{res}(g,b_k) where g(z) = e^{iaz}f(z)

    Lemma: Let f(z) has a simple pole at c with residue \rho. Let \gamma be the curve \gamma(t) = c + re^{it} \mbox{ for }a\leq t\leq b.
    Then, \lim_{r\to 0^+} \int_{\gamma}f(z) dz = i \rho (b-a)

    Now we can evaluate,
    \int_{-\infty}^{\infty}\frac{3\sin 2x}{x(x^2+9)} dx

    Consider the meromorphic function,
    f(z) = \frac{3}{z(z^2+9)}
    The bad thing is that we cannot use Jordan's Lemma because the poles of this function are z=0,z=\pm 3i. We disregard z=-3i because we never will be working in the lower half plane, however z=0 lies on the real axis! So the conditions of Jordan's Lemma are not fullfiled. In order to avoid that we will use the other lemma by drawing a small circular contor around is "bad" point. Look at picture below.

    The bottom is divided into three parts: \gamma_1 which is the left-line segment moving from -\infty to -r. Also, \gamma_2 which is the semi-circular contour (notice, this is negatively oriented). Finally, \gamma_3 which is moving from r to \infty.

    Now the only pole in upper half plane of g(z) = \frac{3e^{2zi}}{z(z^2+9)} is z=3i. And \mbox{res}(g,3i)=\lim_{z\to 3i} (z-3i)\cdot \frac{3e^{2zi}}{(z-3i)(z+3i)} = \frac{3e^{-6}}{6i} = - \frac{1}{2}ie^{-6}

    Now by this generalization of Michael Jordan's Lemma we have (notice the negative):
    \int_{-\infty}^{-r}\frac{3e^{2xi}}{x(x^2+9)} dx - \int_{\gamma_2} \frac{3e^{2zi}}{z(z^2+9)} + \int_r^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx = 2\pi i \left( -\frac{1}{2}ie^{-6} \right) = \pi e^{-6}

    Now if we take the limit at r\to 0^+ by that Lemma we have:
    \int_{-\infty}^{0^-} \frac{3e^{2xi}}{x(x^2+9)}dx - \mbox{res}(g,0)\cdot (\pi - 0)\cdot i + \int_{0^+}^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx = \pi e^{-6}

    Since \mbox{res}(g,0) = \lim_{z\to 0} z\cdot \frac{e^{2zi}}{z(z^2+9)} = \frac{1}{9}

    We have,
    \int_{-\infty}^{0^-} \frac{3e^{2xi}}{x(x^2+9)} dx + \int_{0^+}^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx = \pi e^{-6}+ \frac{1}{9}\pi i

    Thus,
    \int_{-\infty}^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx= \int_{-\infty}^{\infty} \frac{3\cos 2x}{x(x^2+9)} + i \int_{-\infty}^{\infty} \frac{3\sin 2x}{x(x^2+9)} dx = \pi e^{-6}+\frac{1}{9}\pi i

    Thus,
    \int_{-\infty}^{\infty} \frac{3\cos 2x}{x(x^2+9)} \ dx = \pi e^{-6}
    And,
    \int_{-\infty}^{\infty} \frac{3\sin 2x}{x(x^2+9)} \ dx = \frac{1}{9}\pi
    Attached Thumbnails Attached Thumbnails Complex Analysis - Integration-picture19.gif  
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