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Math Help - How to to integral - R in C

  1. #1
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    How to to integral - R in C

    Often when I solve a real integral in the complex plane I take a half circle in the upper plane an calculate the residues to get the integral. But what if theres a singular point in origo?

    Here's an example of what I mean. \displaystyle \int_{- \infty}^\infty \frac{\sin x}{x(x-i)(x+i)} dx

    If anyone could show a method to solve this kind of integral I go.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    On the half circunferemce (radius R) you mentioned, find:

    \displaystyle\int_{C}\dfrac{e^{iz}dz}{z(z^2+1)}

    surrounding the origin with a little circumference of radius \epsilon in the upper plane. After, take limits \epsilon\rightarrow{0^+},\;R\rightarrow{+\infty}.

    Regards.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Try with the closed 'red path' in figure...



    ... computing the limit of the Cauchy integral when r \rightarrow 0 and R \rightarrow \infty...

    Kind regards

    \chi \sigma
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    Notice, by the way, you will be integrating around the small circle clockwise rather than counter-clockwise. Be careful of the sign.
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    Before I get my hands dirty. Even if it's not the primary issue, I rewrote the integral like this... \displaystyle \frac{1}{2i} \int \dfrac{e^{iz}dz}{z(z-i)(z+i)} - \frac{1}{2i} \int \dfrac{e^{-iz}dz}{z(z-i)(z+i)}

    Now if I try to find the residues... How would I go about it. If I calculate the residue in origo in the upper plane, what about the lower plane..?



    Should I just calculate all residues clockwise in the inner circle..? Something like this. \displaystyle \Res [f(z) ; z = i, z = 0, z = -i ] where f(z) is given by the integral.
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  6. #6
    MHF Contributor chisigma's Avatar
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    For 'semplicity's sake' as first step we can integrate the function \displaystyle f(z)= \frac{\sin z}{z\ (z^{2}+1)} 'counterclockwise' along the 'red path' of the figure...



    Is...

    \displaystyle \int_{c} f(z)\ dz = 2\ \pi\ i\ \sum_{k} r_{k} (1)

    ... where the r_{k} are the residues of f(z) inside ther red path. How many poles of f(*) there are inside the 'red path'?... of course the only pole is at z=+i so that is...

    \displaystyle \int_{c} f(z)\ dz = 2\ \pi\ i\ \lim_{z \rightarrow +i} \frac{\sin z}{z\ (z+i)}= \frac{\pi}{2}\ (e-\frac{1}{e}) (2)

    Now what is the next step?...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    Now what is the next step?...
    Yeah that's the money question! Something like this..?

    \displaystyle -\int_{C_r} f(z) dz = - 2\pi i Res \left[ \frac{\sin z}{z^2+1} \right]_{z = 0}
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  8. #8
    MHF Contributor chisigma's Avatar
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    The second step is quite easy: You observe [always looking at the figure...] that is...

    \displaystyle \int_{c} f(z)\ dz = \int_{-R}^{-r} f(z)\ dz + \int_{-r}^{+r} f(z)\ dz + \int_{+r}^{+R} f(z)\ dz + \int_{+R}^{-R} f(z)\ dz = \frac{\pi}{2}\ (e-\frac{1}{e}) (1)

    ... so that the next step is [logically] the computation of the integrals from -r to +r and from +R to -R... how to do?...

    Kind regards

    \chi \sigma
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  9. #9
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    Quote Originally Posted by chisigma View Post
    how to do?...
    I'm not quite sure of why you're asking me that..? My bets are that you have a hunch, since you're stating that it's 'quite easy'. Ah well, maybe it's an linguistics issue..?

    The point of this thread was to learn how to calculate integrals in the complex plane where theres a singularity in origo. If anyone would help to or show me how to do that I'd be thrilled.
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  10. #10
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    Hi again! I was shown a how to solve exercises like this some weeks ago. Now I'm closing in on the exam and when I try to work through the solutions I get somewhat puzzled.

    \displaystyle \int_{- \infty}^\infty \frac{\sin x}{x(x-i)(x+2i)} dx

    First step was to use Eulers formulas, in particular sin x.

    \displaystyle \frac{1}{2i}\int_{- \infty}^\infty \frac{e^{ix}}{x(x-i)(x+2i)} dx - \frac{1}{2i}\int_{- \infty}^\infty \frac{e^{-ix}}{x(x-i)(x+2i)} dx

    Now he calculated the residues of each term by it self.

    2 \pi i Res[f(z); z=0, z=i] where f(z) i given by \displaystyle f(z) = \frac{1}{2i}\int_{- \infty}^\infty \frac{e^{iz}}{z(z-i)(z+2i)} dz

    2 \pi i Res[g(z); z=0, z=-2i] where g(z) i given by \displaystyle g(z) =  \frac{1}{2i}\int_{- \infty}^\infty \frac{e^{-iz}}{z(z-i)(z+2i)} dz

    I know how to calculate the residues, but I don't understand why z = 0 and z = i goes to f(z). Or why z = 0 and z = -2i goes to g(z)..? Feels like I'm missing the main issue here... Any help would be appreciated!
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