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Help with an irregular integral

  1. Dec 9, 2012 #1
    I am looking for help with doing the following integral :
    [tex]\frac{1}{2\pi i}\int_{1}^{\infty}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} \right )\frac{dx}{x\left(\ln x+z\right)}\;\;\;\;z\in \mathbb{C}[/tex]
    i tried to transform it into a complex integral along a 'keyhole' contour, with a branch cut along the +ive real line [itex]\left[1,\infty\right)[/itex]. but then [itex]\;\ln x\;[/itex] would be transformed into [itex]\;\ln x+2\pi i\;[/itex] when doing the integral along [itex]\left(\infty,1\right]\;[/itex] which doesn't add up nicely to the portion along [itex]\left[1,\infty\right)[/itex]. any insights are appreciated.
     
  2. jcsd
  3. Dec 9, 2012 #2

    haruspex

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    Doesn't this simplify fairly easily?
     
  4. Dec 10, 2012 #3
    you would think !!! but no, it doesn't ....
     
  5. Dec 10, 2012 #4
    The integral above is equivalent to :

    [tex] \int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx [/tex]
    And
    [tex]\int_{1}^{\infty}\sum_{n=1}^{\infty}\frac{\sin(2 \pi n x)}{n\pi}\left(\frac{1}{x\left(\ln x+z\right)}\right)dx [/tex]
     
  6. Dec 10, 2012 #5

    haruspex

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    [itex]\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} = \frac{e^{-2\pi i x}(e^{2\pi i x}-1)}{1-e^{2\pi i x}} = -e^{-2\pi i x}[/itex]
    I end up with [itex]\int_0^∞\frac{1-2e^u}{2(u+z)}du[/itex]
    which surely doesn't converge?
     
  7. Dec 12, 2012 #6
    you missed the fact that the inverse of the complex exponential - the complex [itex]\log[/itex] function- is multivalued. namely :
    [tex] \frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}=-e^{-2\pi i x}=e^{-2\pi i \left(x-1/2\right)}=e^{-2\pi i \left(\left \{ x \right \}-1/2\right)}[/tex]
    Where [itex] \left \{ x \right \}[/itex] is the fractional part of x. Thus:
    [tex] \frac{1}{2\pi i}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}\right)=\frac{1}{2}-\left \{ x \right \}[/tex]
    Another way to think of it is to take the Taylor expansion of the [itex] \log[/itex]:
    [tex] \frac{1}{2\pi i}\left(\ln\left(1-e^{-2\pi i x}\right)-\ln(1-e^{2\pi i x})\right)=\frac{1}{2\pi i}\sum_{k=1}^{\infty}\frac{e^{2\pi i k x}-e^{-2\pi i kx}}{k}=\sum_{n=1}^{\infty}\frac{\sin(2\pi k x)}{k\pi}[/tex]
    Which in turn is the Fourier expansion of [itex]\frac{1}{2}-\left \{ x \right \} [/itex]
    Using these facts, we can prove that the integral in question is equal to the limit:
    [tex] e^{-z}\text{Ei}(z)+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left(n+\frac{1}{2} \right )\ln\left(\frac{\ln(n+1)+z}{\ln(n)+z} \right )-e^{-z}\text{Ei}(z+\ln N)[/tex]
    Where [itex] \text{Ei}(z)[/itex] is the exponential integral function. But i'm stuck with this cumbersome limit
     
    Last edited: Dec 12, 2012
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