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Math Help - Calculating residues to solve an integral

  1. #1
    MHF Contributor arbolis's Avatar
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    Calculating residues to solve an integral

    The integral is I=\int _0 ^{+\infty} \frac{x \sin (x)}{x^2+2}dx. I'll probably choose a semi circle as contour since the poles aren't on the real axis. Furthermore I think that I=\frac{1}{2} \int _{-\infty}^{+\infty } \frac{x \sin (x)}{x^2+2}dx.
    Let f(z)=\frac{z \sin z}{z^2+2}. The singularities of f are when z=\pm i \sqrt 2.
    Now I want to calculate the residue of f at say one singularity. But it gives infinity I think. So I believe it's a pole of order greater than 1 but I'm unsure. I tried to use the Taylor series of sin(z) in order to see what happens in z=i \sqrt 2 but I didn't reach anything.
    So I don't know how to go further.
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  2. #2
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    There shouldn't be a problem there. Can you show me your calculation of the residue of f?
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Generally for integrals like this involving  \sin(z) or  \cos(z) , you'll need to switch them out for  e^{iz} . Otherwise the integrals over parts of your contour might not go to zero.

    Therefore I would integrate  \displaystyle f(z)=\frac{z e^{iz}}{z^2+2} .

    Note that  \displaystyle \int _{-\infty}^{\infty } \frac{x\sin (x)}{x^2+2}dx = \text{Im}\left(\int _{-\infty}^{\infty } \frac{ze^{iz}}{z^2+2}dz\right) .
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    MHF Contributor arbolis's Avatar
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    Good idea chip. However I didn't have more chance.
    Res(f,i \sqrt 2 ) = \lim _{z \to i \sqrt 2} \frac{z^2e^{iz}}{z^2+2}=\frac{e^{iz}}{1+\frac{2}{z  ^2}}. The numerator tends to e^{-i \sqrt 2} while the denominator goes to 0 which is really dramatic. It happens whether or not I choose e^{iz} or \sin (z)=\frac{e^{iz} - e^{-iz}}{2i} since the denominator doesn't change.
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  5. #5
    Behold, the power of SARDINES!
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    \displaystyle f(z)=\frac{ze^{iz}}{z^2+2}=\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}

    Now take the limit

    \displaystyle \lim_{z \to i\sqrt{2}}(z-i\sqrt{2})\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}=\lim_{z \to i\sqrt{2}}\frac{ze^{-iz}}{(z+i\sqrt{2})}=\frac{i\sqrt{2}e^{\sqrt{2}}}{i  2\sqrt{2}}=\frac{e^{\sqrt{2}}}{2}
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  6. #6
    MHF Contributor arbolis's Avatar
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    Thanks TheEmptySet! I missed that despite you've done the same yesterday!
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by TheEmptySet View Post
    \displaystyle f(z)=\frac{ze^{iz}}{z^2+2}=\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}

    Now take the limit

    \displaystyle \lim_{z \to i\sqrt{2}}(z-i\sqrt{2})\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}=\lim_{z \to i\sqrt{2}}\frac{ze^{-iz}}{(z+i\sqrt{2})}=\frac{i\sqrt{2}e^{\sqrt{2}}}{i  2\sqrt{2}}=\frac{e^{\sqrt{2}}}{2}
    I think the residue should be  \displaystyle \frac{e^{-\sqrt{2}}}{2} .

    You accidently had  \displaystyle e^{-iz} instead of  \displaystyle e^{iz} .
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  8. #8
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I think the residue should be  \displaystyle \frac{e^{-\sqrt{2}}}{2} .

    You accidently had  \displaystyle e^{-iz} instead of  \displaystyle e^{iz} .
    I was about to ask the same. I get the residue = \frac{e^{-\sqrt{2}}}{2}.
    Edit: I get the answer now. I hope my justification is good enough. (it's very similar to the one of Chip in a previous thread).
    Last edited by arbolis; August 3rd 2010 at 06:50 PM.
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