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Math Help - Complex integration

  1. #1
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    Complex integration

    Hi,

    while trying to evaluate below

    <br />
\int_{-\infty}^{\infty} \! \frac{\sin{x}}{x} \, dx<br />

    I was hoping to use
    1. Use complex numbers i.e pole at x=0
    <br />
\int_{-\infty}^{\infty} \! f(x) \, dx  = 2\pi\, i \sum_{res\, upper\, hp} {f(x)} \, + \pi\, i \sum_{res\, real\, axis} {f(x)}<br />
    which gives 0

    2. Expand by sin(x) by Taylor series around 0 and multiply by x
    this gives a divergent series

    which one is correct?

    thanks
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  2. #2
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    Quote Originally Posted by krindik View Post
    Hi,

    while trying to evaluate below

    <br />
\int_{-\infty}^{\infty} \! \frac{\sin{x}}{x} \, dx<br />

    I was hoping to use
    1. Use complex numbers i.e pole at x=0
    <br />
\int_{-\infty}^{\infty} \! f(x) \, dx  = 2\pi\, i \sum_{res\, upper\, hp} {f(x)} \, + \pi\, i \sum_{res\, real\, axis} {f(x)}<br />
    which gives 0

    2. Expand by sin(x) by Taylor series around 0 and multiply by x
    this gives a divergent series

    which one is correct?

    thanks
    neither. \int_{- \infty}^\infty \frac {\sin x}x~dx = \pi

    here is how you do it. note that \int_{- \infty}^\infty \frac {\sin x}x~dx = \mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix}}x~dx

    Because of the pole of the new integrand at x = 0, we instead compute \mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix} - 1}x~dx

    now continue with the integration using a semi-circular contour in the upper half plane, based on the real axis.


    with sum work, i suppose your second method could work. but you multiply by 1/x not x
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  3. #3
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    Thanks.
    However can u pls tell me

    1. Why method 1 is wrong as it is the usual method for integrals where poles are on real axis

    2. As I understand

    <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix}}x~dx<br />

    and

    <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix} - 1}x~dx<br />

    both have poles at x = 0. Why did u take latter? Would u pls elaborate?
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    Quote Originally Posted by krindik View Post
    Thanks.
    However can u pls tell me

    1. Why method 1 is wrong as it is the usual method for integrals where poles are on real axis
    really? and what method is that? it seems like you were trying to apply the residue theorem, but it looks strange to me. are you sure you computed the residues correctly?

    2. As I understand

    <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix}}x~dx<br />

    and

    <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix} - 1}x~dx<br />

    both have poles at x = 0. Why did u take latter? Would u pls elaborate?
    on the contrary, the second integrand has no poles.
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  5. #5
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    really? and what method is that? it seems like you were trying to apply the residue theorem, but it looks strange to me. are you sure you computed the residues correctly?
    yes, I was referring to the Residue theorem.
    Res(sin(x)/x) at x=0 is 0 isn't it?

    <br /> <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix}}x~dx<br />

    and

    <br /> <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix} - 1}x~dx<br />

    both have poles at x=0. Could u pls explain with some basic principles why it is not?

    thanks
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  6. #6
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    Quote Originally Posted by krindik View Post
    yes, I was referring to the Residue theorem.
    Res(sin(x)/x) at x=0 is 0 isn't it?
    yes, it is 0. but the conditions to apply the residue theorem to this function directly are not fulfilled. that's why we needed to tweak things.

    <br /> <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix}}x~dx<br />

    and

    <br /> <br />
\mathfrak{Im} \int_{- \infty}^\infty \frac {e^{ix} - 1}x~dx<br />

    both have poles at x=0. Could u pls explain with some basic principles why it is not?

    thanks
    the second has a removable singularity at x = 0, not a pole. look up what it means to have a pole and a removable singularity. you will see that the two integrals are different in that respect.
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  7. #7
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    Thank you very much for pointing me in the correct direction.


    To apply the Residue theorem can't we create a contour like the one in the image?
    File:Contour of KKR.svg - Wikipedia, the free encyclopedia
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  8. #8
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    Quote Originally Posted by krindik View Post
    Thank you very much for pointing me in the correct direction.


    To apply the Residue theorem can't we create a contour like the one in the image?
    File:Contour of KKR.svg - Wikipedia, the free encyclopedia
    perhaps. but from the little attention that i paid in complex analysis class, i picked up that with integrals of the form \int_{- \infty}^\infty \frac {P(x)}{Q(x)} \sin x~dx or \int_{- \infty}^\infty \frac {P(x)}{Q(x)} \cos x~dx, you use the semicircular contour i mentioned before. for this particular guy, the set up i did is the way to go using contour integration.
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