Pardon my ignorance. I don't know enough calculus to understand this. I assume this is a very easy question for this amazing site. I saw this on the The Theory of Riemann Zeta Function Book.
$$\sum_{n=2}^{\infty}\pi (n)\int_{n}^{n+1}\frac{s}{x(x^{s}-1)}dx =s\int_{2}^{\infty}\frac{\pi (x)}{x(x^{s}-1)}dx$$
where $\pi(n)$ is the prime counting function.
Can someone explain to me how did that sum symbol get lost?
First, remember that if $a<b<c$, then $\int_{a}^{c}f\left(x\right)\;dx=\int_{a}^{b}f\left(x\right)\;dx+\int_{b}^{c}f\left(x\right)\;dx$. Also, since $\pi$ is constant on $\left[n,n+1\right)$, you can push it into the integral as $\pi\left(x\right)$. Then, if you look at the partial sums, at the $n$th term, you have
$$\pi\left(2\right)\int_{2}^{3}\frac{s}{x\left(x^{s}-1\right)}dx+\pi\left(3\right)\int_{3}^{4}\frac{s}{x\left(x^{s}-1\right)}dx+\cdots+\pi\left(n\right)\int_{n}^{n+1}\frac{s}{x\left(x^{s}-1\right)}dx=\\s\int_{2}^{3}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx+s\int_{3}^{4}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx+\cdots+s\int_{n}^{n+1}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx=\\s\int_{2}^{n+1}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx.$$
The sum is just $$\lim_{n\to\infty}s\int_{2}^{n+1}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx=s\int_{2}^{\infty}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx.$$
Linearity of integral and $\pi$ is constant $\pi (n)$ on the interval $[n, n+1)$.
EDIT: Also: the set $[2, \infty)$ is broken to intervals $[n, n+1],$ that's not called linearity... (is it additivity?)
