# Certain Complex Integral

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?

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.

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

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:

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

• Is there a way to calculate it without using wolfram alpha? Oct 12, 2018 at 2:03
• @KentaS see my latest edit. Oct 12, 2018 at 14:06