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Math Help - Problem With Residues

  1. #1
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    Problem With Residues

    hi all, my first post; had a minor headache with this problem lol.

    PROBLEM 1:
    Finding Residue:
    -----------------
    find Res(g,0) for  g(z) = z^{-2}coshz

    My Attempt/Solution:
    -----------------
    I know  coshz = 1 + \frac{x^2}{2!} + \frac{z^4}{4!} ....

    so now  z^{-2}coshz = z^{-2} (1 + \frac{z^2}{2!} + \frac{z^4}{4!} ... ) = \frac{1}{z^2} + \frac{z^2}{2!z^{2}} + \frac{z^4}{4!z^{2}} ... = \frac{1}{z^2} + \frac{1}{2!} + \frac{z^2}{4!}

    we know the residue is the coefficient of the -1th term (or the coefficient of z^{-1}) but there is no -1th term as you can see. So does that mean the residue is 0, or am I missing something?


    PROBLEM 2:
    Finding Integral:
    ----------------------------
    Evaluate by integrating around a suitable closed contour:
    \int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4}

    My Attempt/Solution:
    -----------------
    First consider the intergral,
     I_R = \int_{\gamma^R} \frac{e^{3iz}}{(z+2i)(z-2i)}dz where  \gamma^R = \gamma^R_1 + \gamma^R_2 and \gamma^R_1 = \{|z| = R , Im(z) > 0\} and \gamma^R_2 = \{z = Re(z) , -R\leq z \leq R\}

    Now, let g(z) = \frac{e^{3iz}}{(z+2i)(z-2i)}. The integrand has 2 simple poles, one at 2i and one at -2i. Only 2i is inside the contour so:

    Res(g,2i) = \lim_{z \to 2i}\frac{(z-2i)e^{3iz}}{(z+2i)(z-2i)} = \lim_{z \to 2i}\frac{e^{3iz}}{(z+2i)} = \frac{e^{-6}}{4i}

    By Cauchy's Residue Theorem, we have:
     I_R = (2\pi i)\frac{e^{-6}}{4i} = \frac{\pi e^{-6}}{2}

    To show that \int_{\gamma^R_1}g(z)dz \to 0, R \to \infty, we apply Jordan's Lemma : M(R) \leq \frac{1}{R^2-4}.

    So now this means,  \int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4} = \frac{\pi e^{-6}}{2}

    Please could you verify whether I have worked these out correctly. Thanks a great deal, highly appreciated .
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  2. #2
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    Quote Originally Posted by mathfied View Post
    find Res(g,0) for  g(z) = z^{-2}coshz
    \cosh z = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} thus, z^{-2}\cosh z = \sum_{n=0}^{\infty} \frac{z^{2n-2}}{(2n)!}.
    Since 2n-2 is always even it means there is no z^{-1} term and so the residue is zero.
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  3. #3
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    Quote Originally Posted by mathfied View Post
    Evaluate by integrating around a suitable closed contour:
    \int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4}
    Let f(z) = \frac{e^{3iz}}{z^2+4}. By using a semi-circle contour with R>2 we find:
    \int_{-R}^R \frac{e^{3ix}}{x^2+4} dx + \int_0^{\pi} \frac{Rie^{i\theta}e^{iRe^{i\theta}}}{R^2e^{2i\the  ta}+4}d\theta = \oint_{\Gamma} \frac{e^{3iz}}{z^2+4} dz.
    Now take limit R\to \infty, the middle integral is zero (by Jordan's lemma) while the RHS integral is computed by residue theorem:
    \int_{-\infty}^{\infty} \frac{e^{3ix}}{x^2+4}dx = 2\pi i\cdot \mbox{res}(f,2i).
    It remains to compute \mbox{res}(f,2i) note that this is a simple pole:
    \mbox{res}(f,2i) = \lim_{z\to 2i} (z-2i)\cdot \frac{e^{3iz}}{(z+2i)(z-2i)} = \frac{e^{-6}}{4i}.
    Thus,
    \int_{-\infty}^{\infty} \frac{e^{3ix}}{x^2+4}dx = 2\pi i \cdot \frac{e^{-6}}{4i} = \frac{\pi}{2e^6}.
    This mean by equating real and imaginary parts,
    \int_{-\infty}^{\infty}\frac{\cos 3x}{x^2+4} dx = \frac{\pi}{2e^6}.
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  4. #4
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    Great! So my method was correct. Thanks for the help. Very much appreciated!
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