I'm trying to compute the following integral ($ \vartheta(x) = \sum\limits_{p \leq x}\log(p) $)

$$\int\limits_{0}^{\infty}\vartheta(e^x) e^{-(1+s)x} \text{dx}$$

The result is supposed to be $ \frac{\sum\log(p)/p^{s+1}}{1+s} $, but I'm having trouble doing it. I tried to substitute $ t=e^x, dt/t=dx $ and I got

$$\int\limits_{1}^{\infty} \vartheta(t)t^{-(2+s)} \text{dx}$$

And I'm stuck. I don't know how to evaluate an integral with a sum inside that depends on in


Change the order of summation and integration. Since everything is non-negative, that is unproblematic.

\begin{align} \int_0^\infty \biggl(\sum_{p\leqslant e^x} \log p\biggr)e^{-(1+s)x}\,dx &= \int_0^\infty e^{-(1+s)x} \sum_p [p \leqslant e^x]\log p\,dx\\ &= \sum_p \log p\cdot\int_0^\infty e^{-(1+s)x}[p \leqslant e^x]\,dx\\ &= \sum_p \log p\cdot\int_0^\infty e^{-(1+s)x} [x \geqslant \log p]\,dx\\ &= \sum_p \log p\cdot\int_{\log p}^\infty e^{-(1+s)x}\,dx\\ &= \sum_p \log p\cdot \frac{e^{-(1+s)\log p}}{(1+s)}\\ &= \frac{1}{1+s}\sum_p \frac{\log p}{p^{1+s}}, \end{align}

where $[\,\cdot\,]$ is the Iverson bracket, $[x \geqslant \log p]$ is $1$ if $x\geqslant \log p$, and $0$ if $x < \log p$.

  • $\begingroup$ Thank you again, I must have forgotten the time I read Knuth's book who used this bracket crazily often $\endgroup$ – Jytug Nov 30 '14 at 20:25

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