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Thread: Residue Theorem

  1. #1
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    Residue Theorem

    So I'm working on a Residue theory problem set. I was doing fine until I got to this one:

    The integral, from -infinity to infinity of:
    (cosx-isinx)/(1+x^2) dx

    My first idea was to rewrite the top by adding adding and subtracting cosx:
    (-cosx-isinx)+2cosx=-e^(ix)+2cosx

    And then proceed by splitting up into 2 integrals.

    int(2cosx/(1+x^2))-int(e^ix/(1+x^2))

    But I am stuck on this. The first integral shouldn't be too bad. I think there are 2 simple poles, at +/- i. I would integrate around the upper half disk including i. so the integral would just be 2pi(Res(f,i)+Res(f,-i)).

    For the second one I want to proceed the same way, but I shaky with figuring out poles with something new (e^ix) in the numerator.
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  2. #2
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    Integrate $\displaystyle \frac{e^{iz}}{1+z^2}$ round the usual contour (real axis + upper semicircle). That will give you the value of $\displaystyle \int_{-\infty}^{\infty}\frac{\cos x + i\sin x}{1+x^2}dx$. Then take the complex conjugate to get the integral that you want. But in fact the answer should turn out to be real anyway, because the sin(x) part of the integrand is an odd function, so its integral should be 0.
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by Opalg View Post
    Integrate $\displaystyle \frac{e^{iz}}{1+z^2}$ round the usual contour (real axis + upper semicircle). That will give you the value of $\displaystyle \int_{-\infty}^{\infty}\frac{\cos x + i\sin x}{1+x^2}dx$. Then take the complex conjugate to get the integral that you want. But in fact the answer should turn out to be real anyway, because the sin(x) part of the integrand is an odd function, so its integral should be 0.
    Just a question ^^'

    Why don't we take $\displaystyle e^{-ix}=\cos(x)-i\sin(x)$ ?
    Is it because the integral wouldn't go to 0 on the semi-circle ?

    Just want to make sure
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    Quote Originally Posted by Opalg View Post
    Integrate $\displaystyle \frac{e^{iz}}{1+z^2}$ round the usual contour (real axis + upper semicircle). That will give you the value of $\displaystyle \int_{-\infty}^{\infty}\frac{\cos x + i\sin x}{1+x^2}dx$. Then take the complex conjugate to get the integral that you want. But in fact the answer should turn out to be real anyway, because the sin(x) part of the integrand is an odd function, so its integral should be 0.
    Great idea! Can you typically take a conjugate, integrate, and then take the conjugate again to get an integral?
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    Just a question ^^'

    Why don't we take $\displaystyle e^{-ix}=\cos(x)-i\sin(x)$ ?
    Is it because the integral wouldn't go to 0 on the semi-circle ?

    Just want to make sure
    Yes, come to think of it that would be just as good as my method. But you would then have to go round the lower semicircle, because $\displaystyle e^{-i(x+iy)} = e^ye^{-ix}$. This gets large when y is positive, but stays small if y is negative. So if you do the question this way then you would have to use the residue at i instead of +i.
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  6. #6
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    Quote Originally Posted by robeuler View Post
    Can you typically take a conjugate, integrate, and then take the conjugate again to get an integral?
    Yes, if it's an integral along the real axis. It wouldn't work for a general contour integral. But in this case the integral around the semicircular part of the contour goes to zero as the radius goes to +∞, so it's OK.
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