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I was trying to generalize the Riemann's prime number formula for $\pi(x)$ to a general algebraic field $K$, and came across the integral:

$$f(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{d}{ds}\left[\frac{\log s}{s}\right]x^s ds (x>1, c>0).$$

I have proved through simple calculation that this integral converges and for $\forall \epsilon>0$, $|f(x)|=O(x^\epsilon)$. Can anybody help me, please?

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We can express your integral as an inverse Mellin transform or inverse Laplace Transform:

$$ f(x) = \mathcal{M}^{-1}\left[\frac{1}{s^2}-\frac{\log (s)}{s^2}\right](1/x) = \mathcal{L}^{-1}\left[\frac{1}{s^2}-\frac{\log (s)}{s^2}\right](\log x). $$

The latter can be calculated here.

This leads to

$$ f(x) = \log (x) (\log (\log (x))+\gamma ), $$

where $\gamma$ is the Euler Mascheroni constant.


Edit

Here's a way to compute the inverse Laplace transform:

  • Integrate by parts
  • Introduce a parameter $a = -1$
  • Differentiate and integrate w.r.t. $a$ and use the linearity of the transform
  • Integrate
  • Evaluate the transform
  • Differentiate
  • Substitute $a = -1$

$$ \begin{align*} \mathcal{L}_{s}^{-1}\left[\frac{d}{ds} \frac{\log (s)}{s}\right](t) &= -t \, \mathcal{L}_{s}^{-1}\left[\frac{\log (s)}{s}\right](t) \\ &= -t \, \mathcal{L}_{s}^{-1}\left[\log (s) \, s^a\right](t) \\ &= -t \, \frac{d}{da} \mathcal{L}_{s}^{-1}\left[\int \log (s) \, s^a da\right](t) \\ &= -t \, \frac{d}{da} \mathcal{L}_{s}^{-1}\left[s^a\right](t) \\ &= -t \, \frac{d}{da} \frac{t^{-a-1}}{\Gamma(-a)} \\ &= \frac{t^{-a} (\Gamma (-a) \log (t)-\Gamma'(-a))}{\Gamma (-a)^2} \\ &= t (\log (t) - \Gamma'(1)) = t (\log (t) + \gamma). \end{align*} $$

Generically (ignoring convergence, etc) we can generalize:

$$ \mathcal{L}_{s}^{-1}[f(s)\log(s)](t) = \left.\frac{d}{da}\right|_{a=0} \mathcal{L}_{s}^{-1}[f(s)s^a](t). $$

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  • $\begingroup$ Is there a way to calculate it without using wolfram alpha? $\endgroup$
    – Kenta S
    Oct 12, 2018 at 2:03
  • $\begingroup$ @KentaS see my latest edit. $\endgroup$
    – Greg Hurst
    Oct 12, 2018 at 14:06

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