Chebyshev’s $\theta_1(x) = \sum_{n \leqslant x } (x-n) \Lambda_1 (n)$

Question

Let $\theta_1 = \int_{1}^{x}\theta (t)dt$, for $x \gt 1$ where $\theta(x)$ is the Chebyshev’s function.

Letting $\Lambda_1(n) = \log n,\;$ if $n$ is prime, then $\Lambda_1(n)=0.$ Otherwise prove that:

$$\theta_1(x) = \sum_{n \leqslant x } (x-n) \Lambda_1 (n).$$

Answer 1

Method 1: Geometric interpretation

From single-variable calculus, integrals can be interpreted as area under the curve. To calculate the area under the step-function's "curve," we consider calculating them vertically. Since each rectangle has area $(x-n)\Lambda_1(n)$, we conclude

$$ \vartheta_1(x)=\sum_{n\le x}(x-n)\Lambda_1(n) $$

Method 2: Abel's summation (aka Riemann-Stieltjes integration)

Consider reversing Abel's summation:

$$ \begin{aligned} \vartheta_1(x) &=\int_1^x\vartheta(t)\mathrm dt =t\vartheta(t)|_1^x-\int_1^x t\mathrm d\vartheta(t) \\ &=x\vartheta(x)-\sum_{1<n\le x}n\Lambda_1(n) \end{aligned} $$

Since $\Lambda_1(1)=0$, it does not make any difference if we change the summation interval to $1\le n\le x$. As a result, we get

$$ \vartheta_1(x)=x\sum_{n\le x}\Lambda_1(n)-\sum_{n\le x}n\Lambda_1(n)=\sum_{n\le x}(x-n)\Lambda_1(n) $$

Answer 2

$$\theta_1(x) = \int_1^x \theta(t) dt= \int_1^x \left(\sum_{n\le t} \Lambda_1(n) \right) dt =\sum_{n\le x} \int_{n}^x \Lambda_1(n) dt = \sum_{n \le x}(x-n)\Lambda_1(n).$$