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Does this integral converge ?

  1. Feb 1, 2013 #1
    Does this integral converge !?

    [tex] \int_{0}^{\infty}x\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )dx[/tex]
    Where [itex]\zeta(x)[/itex] is the Riemann zeta function. Of course, we can integrate by parts to obtain :

    [tex]\frac{1}{2}-\int_{0}^{\infty}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1}\right) dx [/tex]

    Unfortunately, i don't know how to do this integral, and i don't have mathematica, hence the thread !!
  2. jcsd
  3. Feb 1, 2013 #2


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  4. Feb 1, 2013 #3
    Re: Does this integral converge !?

    And it doesn't converge . i kinda knew it , thanks . i was trying to evaluate a limiting case of the integral :
    [tex] \int_{0}^{\infty}\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )\ln\left(-e^{-zx}\right)dx [/tex]
    For [itex] z\in \mathbb{C} [/itex]
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