Writing Zeta Function In Terms Of The J-Function I am reading through John Derbyshire's "Prime Obsession" and am struggling to understand his argument for why $\frac{1}{s} \log{\zeta(s)}=\int_{0}^{\infty} J(x)x^{-s-1}dx$ where $J(x)$ is defined as $\pi(x)+\frac{1}{2}\pi(\sqrt{x})+\frac{1}{3}\pi(\sqrt[3]{x})+\frac{1}{4}\pi(\sqrt[4]{x})+\frac{1}{5}\pi(\sqrt[5]{x})+...$
Here is what I am getting so far:

*

*I know $\zeta(s)={ \prod_{p} \left(1-p^{-s}\right)^{-1}}$.

*Taking the logarithm, $\log\left(\zeta(s)\right)=-\log(1-\frac{1}{2^s})-\log(1-\frac{1}{3^s})-\log(1-\frac{1}{5^s})+...$

*Recall $S=\sum_{k=0}^{n-1}a\cdot r^k=\frac{1}{1-r}$ whenever $a=1$ and $r\in(-1,1)$. Taking the integral, we have $\int{\frac{1}{1-r}}=\int{1+r+r^2+r^3+...}$, and $-\log(1-r)=r+\frac{r^2}{2}+\frac{r^3}{3}+\frac{r^4}{4}+...$. Then since $0 < \lvert \frac{1}{p^s} \rvert<1$, we can write each term in Euler's product formula as an infinite sum. For example, $-\log(1-\frac{1}{2^s})=\frac{1}{2^s}+\left(\frac{1}{2}\cdot\left(\frac{1}{2^s}\right)^2\right)+\left(\frac{1}{3}\cdot\left(\frac{1}{2^s}\right)^3\right)+\left(\frac{1}{4}\cdot\left(\frac{1}{2^s}\right)^4\right)\dots$

*Any term in this infinite sum of infinite sums can be written as an integral. For example, $\left(\frac{1}{3}\cdot\left(\frac{1}{2^s}\right)^3\right)=\frac{1}{3}\times\frac{1}{2^{3s}}=\frac{1}{3}\cdot{s}\cdot \int_{2^3}^{\infty}x^{-s-1}\: dx$ since $\int_{2^3}^{\infty} x^{-s-1}dx=\left(\frac{1}{s}\cdot\frac{-1}{x^s}\right)\biggr\rvert_{8}^{\infty}=\left(0\right)-\left(\frac{1}{s}\cdot\frac{-1}{8^s}\right)=\frac{1}{s}\times\frac{1}{8^s}$ which is precisely $\frac{s}{3}$ multiples of $\frac{1}{3}\times\frac{1}{2^{3s}}$.

*This is where I am not following. Derbyshire says that this specific term forms a "strip" under the J-Function. Even though the J-Function is a step function, if you think of the integral as area under the curve, the example in the previous step should not be rectangular. Another point that I don't understand is why $\int_{0}^{\infty} J(x)x^{-s-1}dx=\left[\int_{2}^{\infty} \left(\frac{1}{1}\cdot x^{-s-1} dx\right)+\int_{2^2}^{\infty} \left(\frac{1}{2}\cdot x^{-s-1} dx\right)+\int_{2^3}^{\infty} \left(\frac{1}{3}\cdot x^{-s-1} dx\right)+...\right]+\left[\int_{3}^{\infty} \left(\frac{1}{1}\cdot x^{-s-1} dx\right)+\int_{3^2}^{\infty} \left(\frac{1}{2}\cdot x^{-s-1} dx\right)+\int_{3^3}^{\infty} \left(\frac{1}{3}\cdot x^{-s-1} dx\right)+...\right]+\left[\int_{5}^{\infty} \left(\frac{1}{1}\cdot x^{-s-1} dx\right)+\int_{5^2}^{\infty} \left(\frac{1}{2}\cdot x^{-s-1} dx\right)+\int_{5^3}^{\infty} \left(\frac{1}{3}\cdot x^{-s-1} dx\right)+...\right]+...$.

Any insights into this problem?
 A: Let me explain a little of what reuns was intimating in  his\her answer (if it’s still of benefit now). Note that we have
$$\int_n^\infty x^{-s-1}dx=\left.\frac{x^{-s}}{-s}\right|_n^\infty=\frac{1}{s}n^{-s}\,.$$
It then follows that
$$\sum_na_nn^{-s}=s \sum_na_n \int_n^\infty x^{-s-1}dx\,.$$
The next step —which I believe is what you’re interested in understanding—is getting the summation inside the integral. To see that in a step-wise fashion, define the following step-function
$$\chi(n,x):=\left\lbrace
\begin{array}{ll}
1& \mbox{if $n\le x$}\\
0& \mbox{if $n>x$}
\end{array}
\right.$$
then observe that
$$\int_1^\infty\chi(n,x)x^{-s-1}dx= \int_n^\infty\chi(n,x)x^{-s-1}dx+ \int_1^n\chi(n,x)x^{-s-1}dx$$
but the rightmost integral is simply equal to $0$ since $\chi(n,\cdot)$ vanishes on the interval $x\in(1,n)$ and we obtain
$$\int_1^\infty\chi(n,x)x^{-s-1}dx= \int_n^\infty x^{-s-1}dx$$
We can therefore conveniently rewrite our original integral as
$$\sum_na_nn^{-s}=s \sum_na_n \int_1^\infty\chi(n,x)x^{-s-1}dx$$
$$= s \int_1^\infty \sum_na_n  \chi(n,x)x^{-s-1}dx$$
$$= s \int_1^\infty \sum_{n\le x}a_n x^{-s-1}dx\,.$$
We can now apply this identity to the $J$ function; we recall that the $J$ function is the same as
$$J(x)=\sum_{k\ge 1}\sum_{p^k\le x}\frac{1}{k}$$
So from the logarithm of the Riemann function, and using the definition that $a_n=\frac{1}{k}$ if $n=p^k$ in our just-derived identity, we obtain
$$\log\zeta(s)=\sum_{k\ge 1}\sum_{p~\text{prime}}\frac{1}{k}p^{-sk}$$
$$= \sum_{k\ge 1} s \int_1^\infty \sum_{p^k\le x}\frac{1}{k}x^{-s-1}dx$$
$$= s \int_1^\infty \sum_{k\ge 1}\sum_{p^k\le x}\frac{1}{k}x^{-s-1}dx$$
$$=s\int_1^\infty J(x) x^{-s-1}dx\,,$$
as desired.
Hope this helps!
