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Help with a complex integral

  1. May 24, 2012 #1
    greetings , we have the following integral :

    [tex]I(x)=\lim_{T\rightarrow \infty}\frac{1}{2\pi i}\int_{\gamma-iT}^{\gamma+iT}\sin(n\pi s)\frac{x^{s}}{s}ds[/tex]

    n is an integer . and [itex]\gamma >1 [/itex]

    if [itex] x>1[/itex] we can close the contour to the left . namely, consider the contour :

    [tex] C_{a}=C_{1}\cup C_{2}\cup C_{3}\cup C_{4}[/tex]

    where :

    [tex]C_{1}=\left [ \gamma-iT,\gamma+iT \right ] [/tex]

    [tex]C_{2}=\left [ \gamma+iT,-U+iT \right ] [/tex]

    [tex]C_{3}=\left [ -U+iT ,-U-iT \right ] [/tex]

    [tex]C_{4}=\left [ -U-iT ,\gamma-iT \right ] [/tex]

    and [itex]U>>\gamma [/itex]
    then by couchy's theorem :

    [tex] I(x)=I_{1}+I_{2}+I_{3}+I_{4}=0[/tex]

    if [itex] x<1[/itex], we can close the contour to the right via the following contour :

    [tex] C_{b}=C_{1}\cup C_{2}\cup C_{3}\cup C_{4}[/tex]

    [tex]C_{1}=\left [ \gamma-iT,\gamma+iT \right ] [/tex]

    [tex]C_{2}=\left [ \gamma+iT,U+iT \right ] [/tex]

    [tex]C_{3}=\left [ U+iT ,U-iT \right ] [/tex]

    [tex]C_{4}=\left [ U-iT ,\gamma-iT \right ] [/tex]

    then also by couchy's theorem :

    [tex] I(x)=0[/tex]

    the plan is to give an estimate of the integrals along the segments of the rectangular contour, and calculate [itex] I_{1}[/itex] in both cases via the result obtained by cauchy's theorem . however, i don't have the first clue on how to do that, hence the quest !
    Last edited: May 24, 2012
  2. jcsd
  3. May 24, 2012 #2


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    I don't think you mean that. Yes, by Cauchy, [tex] I_{1}+I_{2}+I_{3}+I_{4}=0[/tex], but the challenge is to show that three of these tend to 0 as T tends to infinity, and therefore the fourth (the one that tends to I(x)) does too.
    I got stuck trying to put some bound on sin(s). When s = c+id has a large imaginary component, id, this grows like exp(d). I tried pairing up the negative imaginary points with the positive ones, e.g. doing C3 as the integral from -U to -U+iT of {f(s) - f(complex conjugate of s)}.dt, hoping to get some cancellation. No luck.
  4. May 25, 2012 #3
    yes, sorry .. it was a typo . and i was trying to do exactly the same , no luck :cry:

    remark : the integral of question bears resemblance to the perron's integral, and could be thought of as a laplace inverse of [itex]\frac{\sin(n\pi s)}{s}[/itex], with [itex]\ln x [/itex] as the variable in the 'time' domain.
    Last edited: May 25, 2012
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