Prove that ## g(x)=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)dt ##.

[Math100]

Homework Statement

Let $f(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}1$ and $g(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}$.
a) Find $f(40)$ and $g(40)$.
b) Prove that $g(x)=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)dt$.

Relevant Equations

Abel's summation formula: Suppose $f$ is a function with a continuous derivative on the interval $[x, y]$ where $0<x<y$. Then $\sum_{x\leq n\leq y}a_{n}f(n)=A(y)f(y)-A(x)f(x)-\int_{x}^{y}A(t)f'(t)dt$ with $A(x)=\sum_{0<n\leq x}a_{n}$ where $(a_{n})_{n=0}^{\infty}$ is a sequence of real or complex numbers.

a) $f(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}1=3+13+23=39$
$g(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\log {3}+\log {13}+\log {23}=\log {897}$
b)
Proof:
Let $f(n)=\log {n}$ and $a_{n}=1$ if $n\leq x$ is prime such that $n\equiv 3\pmod {10}$ and $0$ otherwise.
By Abel's summation formula, we have
\begin{align*}
&g(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\sum_{1\leq n\leq x}a_{n}f(n)\\
&=A(x)f(x)-\int_{1}^{x}A(t)f'(t)dt\\
&=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)dt.\\
\end{align*}

 

[fresh_42]

Homework Statement:: Let $f(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}1$ and $g(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}$.
a) Find $f(40)$ and $g(40)$.
b) Prove that $g(x)=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)dt$.
Relevant Equations:: Abel's summation formula: Suppose $f$ is a function with a continuous derivative on the interval $[x, y]$ where $0<x<y$. Then $\sum_{x\leq n\leq y}a_{n}f(n)=A(y)f(y)-A(x)f(x)-\int_{x}^{y}A(t)f'(t)dt$ with $A(x)=\sum_{0<n\leq x}a_{n}$ where $(a_{n})_{n=0}^{\infty}$ is a sequence of real or complex numbers.

a) $f(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}1=3+13+23=39$

$g(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\log {3}+\log {13}+\log {23}=\log {897}$

$\approx 6.8$

b)
Proof:
Let $f(n)=\log {n}$ and $a_{n}=1$ if $n\leq x$ is prime such that $n\equiv 3\pmod {10}$ and $0$ otherwise.
By Abel's summation formula, we have
\begin{align*}
&g(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\sum_{1\leq n\leq x}a_{n}f(n)\\
&=A(x)f(x)-\int_{1}^{x}A(t)f'(t)dt\\
&=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)dt.\\
\end{align*}

It is never a good idea to write two different functions by the same letter. Since $\log (x)$ already has a name, why not use it?
\begin{align*}
&g(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\sum_{2\leq n\leq x}a_{n}\log(n)\\
&=A(x)\log(x)-A(2)\log(2)-\int_{2}^{x}A(t)\log'(t)\, dt\\
&=f(x)\log {x}- \left(\sum_{p\leq 2 \atop{p\equiv 3\pmod{10}}} \right)\cdot \log(2)- \int_{2}^{x} \sum_{n=0}^t \dfrac{a_n}{t}\, d t \\
&=f(x)\log {x}- 0\cdot \log(2) - \int_{2}^x \sum_{p\leq t \atop{p\equiv 3\pmod{10}}} \dfrac{dt}{t}\\
&=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)\, dt.
\end{align*}

 

[Math100]

$\approx 6.8$

It is never a good idea to write two different functions by the same letter. Since $\log (x)$ already has a name, why not use it?
\begin{align*}
&g(x)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\sum_{2\leq n\leq x}a_{n}\log(n)\\
&=A(x)\log(x)-A(2)\log(2)-\int_{2}^{x}A(t)\log'(t)\, dt\\
&=f(x)\log {x}- \left(\sum_{p\leq 2 \atop{p\equiv 3\pmod{10}}} \right)\cdot \log(2)- \int_{2}^{x} \sum_{n=0}^t \dfrac{a_n}{t}\, d t \\
&=f(x)\log {x}- 0\cdot \log(2) - \int_{2}^x \sum_{p\leq t \atop{p\equiv 3\pmod{10}}} \dfrac{dt}{t}\\
&=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)\, dt.
\end{align*}

How did you get $g(40)\approx 6.8$? I thought it's $g(40)\approx 2.95$.

 

[fresh_42]

I just saw that you made a mistake in the first part of the problem.
$$
f(40)=\sum_{p\leq t \atop{p\equiv 3\pmod{10}}}1=\sum_{p\in \{3,13,23\}}1=3
$$
We add $1$ as often as we get summands, not the primes themselves! That would have been $\displaystyle{\sum_{p\leq t \atop{p\equiv 3\pmod{10}}}p}=3+13+23=39.$

How did you get $g(40)\approx 6.8$? I thought it's $g(40)\approx 2.95$.

$g(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\log {3}+\log {13}+\log {23}=\log {897}$

... and $\log 897 = \ln 897= 6.7990558\ldots \approx 6.8$

Let's see if the base was $10$ such that $g(40)=\log_{10}897 \stackrel{?}{\approx} 2.95$ as you think. We have a formula. Abel's summation formula is
$$
\sum_{x<n\leq y}a_n \phi(n)=\left(\sum_{0<n\leq y}a_n\right)\phi(y)-\left(\sum_{0<n\leq x}a_n\right)\phi(x)-\int_x^y\left(\sum_{0<n\leq t}a_n\right)\phi'(t)\,dt
$$
in which we set $\phi(t)=\log(t)$ because $\phi'(t)=\dfrac{d}{dt} \phi(t)=\dfrac{d}{dt} \log(t)=\dfrac{1}{t}.$ Thus
\begin{align*}
g(40)&=f(40)\ln (40)-\int_2^{40}\sum_{p\leq t \atop{p\equiv 3\pmod{10}}} \dfrac{dt}{t}\\
&=3\cdot \ln(40) - \int_2^3 \dfrac{0}{t}\,dt -\int_3^{13}\dfrac{1}{t}\,dt-\int_{13}^{23}\dfrac{2}{t}\,dt-\int_{23}^{40}\dfrac{3}{t}\,dt\\
&=3\cdot \ln(40) -(\ln(13)-\ln(3))-2\cdot (\ln(23)-\ln(13))-3\cdot (\ln(40)-\ln(23))\\
&=\ln(23)+\ln(13)+\ln(3)=\ln (897)\approx 6.8
\end{align*}
... since we used the natural logarithm in Abel's summation formula. We could have used $\log_{10}$ instead, but this would only have added a factor $\ln(10)\approx 2.3$ in every summand and $2.3 \cdot 2.95 \approx 6.8.$

 

[Math100]

I just saw that you made a mistake in the first part of the problem.
$$
f(40)=\sum_{p\leq t \atop{p\equiv 3\pmod{10}}}1=\sum_{p\in \{3,13,23\}}1=3
$$
We add $1$ as often as we get summands, not the primes themselves! That would have been $\displaystyle{\sum_{p\leq t \atop{p\equiv 3\pmod{10}}}p}=3+13+23=39.$



... and $\log 897 = \ln 897= 6.7990558\ldots \approx 6.8$

Let's see if the base was $10$ such that $g(40)=\log_{10}897 \stackrel{?}{\approx} 2.95$ as you think. We have a formula. Abel's summation formula is
$$
\sum_{x<n\leq y}a_n \phi(n)=\left(\sum_{0<n\leq y}a_n\right)\phi(y)-\left(\sum_{0<n\leq x}a_n\right)\phi(x)-\int_x^y\left(\sum_{0<n\leq t}a_n\right)\phi'(t)\,dt
$$
in which we set $\phi(t)=\log(t)$ because $\phi'(t)=\dfrac{d}{dt} \phi(t)=\dfrac{d}{dt} \log(t)=\dfrac{1}{t}.$ Thus
\begin{align*}
g(40)&=f(40)\ln (40)-\int_2^{40}\sum_{p\leq t \atop{p\equiv 3\pmod{10}}} \dfrac{dt}{t}\\
&=3\cdot \ln(40) - \int_2^3 \dfrac{0}{t}\,dt -\int_3^{13}\dfrac{1}{t}\,dt-\int_{13}^{23}\dfrac{2}{t}\,dt-\int_{23}^{40}\dfrac{3}{t}\,dt\\
&=3\cdot \ln(40) -(\ln(13)-\ln(3))-2\cdot (\ln(23)-\ln(13))-3\cdot (\ln(40)-\ln(23))\\
&=\ln(23)+\ln(13)+\ln(3)=\ln (897)\approx 6.8
\end{align*}
... since we used the natural logarithm in Abel's summation formula. We could have used $\log_{10}$ instead, but this would only have added a factor $\ln(10)\approx 2.3$ in every summand and $2.3 \cdot 2.95 \approx 6.8.$

Thank you for pointing that out on part a). Also, another part of this question asks to prove that $g(x)\sim\frac{1}{4}x$ by assuming that $f(x)\sim\frac{1}{4}\pi(x)$. By definitions, both $\pi(x)=\sum_{\substack{prime p\leq x}}1$ and $v(x)=\sum_{\substack{prime p\leq x}}\log {p}$ are step functions for $x\geq 1$. In the process of taking the limit of $\lim_{x\rightarrow \infty}\frac{g(x)}{\frac{1}{4}x}=4\cdot \lim_{x\rightarrow \infty}\frac{g(x)}{x}$, how to prove that $\lim_{x\rightarrow \infty}\frac{g(x)}{x}=\frac{1}{4}$ so that $\lim_{x\rightarrow \infty}\frac{g(x)}{\frac{1}{4}x}=1$?

 

[fresh_42]

Let's use the formulas and https://en.wikipedia.org/wiki/Logarithmic_integral_function.
$$
f(x)\approx \dfrac{|\{p\leq x\, : \,p \text{ prime }\}|}{|\{p\leq x\, : \,p \text{ prime }\wedge p\equiv 3\pmod{10}\}|}\approx\dfrac{\pi(x)}{\varphi(10)}=\dfrac{\pi(x)}{4}\approx \dfrac{x}{4\log(x)}
$$
\begin{align*}
g(x)&=f(x)\log(x) - \int_2^x \dfrac{f(t)}{t}\,dt \approx \dfrac{1}{4}\pi(x)\log(x)- \dfrac{1}{4}\int_2^x \dfrac{\pi(t)}{t}\,dt\\
&\approx \dfrac{x}{4} - \dfrac{1}{4}\int_0^x \dfrac{dt}{\log(t)} + \dfrac{1}{4}\int_0^2 \dfrac{dt}{\log(t)}\approx
\dfrac{x}{4} - \dfrac{1}{4}\underbrace{\operatorname{li}(x)}_{\text{logarithmic integral}} +\dfrac{1}{4}\\
&\approx \dfrac{x+1}{4}-\dfrac{1}{4}\dfrac{x}{\log(x)}\underbrace{\left(1+O(\log^{-1}(x))\right)}_{\stackrel{x\to \infty }{\longrightarrow 1}}\\
&\approx \dfrac{x+1}{4}-\dfrac{\pi(x)}{4}
\end{align*}
Thus
$$
\lim_{x \to \infty}\dfrac{g(x)}{x}=\lim_{x \to \infty}\left(\dfrac{1+(1/x)}{4}-\dfrac{1}{4}\cdot \dfrac{\pi(x)}{x}\right)=\dfrac{1}{4}-\lim_{x \to \infty}\dfrac{1}{4\log(x)}=\dfrac{1}{4}
$$

 

[Math100]

Let's use the formulas and https://en.wikipedia.org/wiki/Logarithmic_integral_function.
$$
f(x)\approx \dfrac{|\{p\leq x\, : \,p \text{ prime }\}|}{|\{p\leq x\, : \,p \text{ prime }\wedge p\equiv 3\pmod{10}\}|}\approx\dfrac{\pi(x)}{\varphi(10)}=\dfrac{\pi(x)}{4}\approx \dfrac{x}{4\log(x)}
$$
\begin{align*}
g(x)&=f(x)\log(x) - \int_2^x \dfrac{f(t)}{t}\,dt \approx \dfrac{1}{4}\pi(x)\log(x)- \dfrac{1}{4}\int_2^x \dfrac{\pi(t)}{t}\,dt\\
&\approx \dfrac{x}{4} - \dfrac{1}{4}\int_0^x \dfrac{dt}{\log(t)} + \dfrac{1}{4}\int_0^2 \dfrac{dt}{\log(t)}\approx
\dfrac{x}{4} - \dfrac{1}{4}\underbrace{\operatorname{li}(x)}_{\text{logarithmic integral}} +\dfrac{1}{4}\\
&\approx \dfrac{x+1}{4}-\dfrac{1}{4}\dfrac{x}{\log(x)}\underbrace{\left(1+O(\log^{-1}(x))\right)}_{\stackrel{x\to \infty }{\longrightarrow 1}}\\
&\approx \dfrac{x+1}{4}-\dfrac{\pi(x)}{4}
\end{align*}
Thus
$$
\lim_{x \to \infty}\dfrac{g(x)}{x}=\lim_{x \to \infty}\left(\dfrac{1+(1/x)}{4}-\dfrac{1}{4}\cdot \dfrac{\pi(x)}{x}\right)=\dfrac{1}{4}-\lim_{x \to \infty}\dfrac{1}{4\log(x)}=\dfrac{1}{4}
$$

So both $\pi(x)\approx \frac{x}{\log {x}}$ and $li(x)\approx \frac{x}{\log {x}}$? How did you get $(1+O(\log^{-1} (x)))$?

 

[fresh_42]

So both $\pi(x)\approx \frac{x}{\log {x}}$ and $li(x)\approx \frac{x}{\log {x}}$?

Roughly. Here are some margins:
https://de.wikipedia.org/wiki/Primzahlsatz#Stärkere_Formen_des_Primzahlsatzes

How did you get $(1+O(\log^{-1} (x)))$?

https://en.wikipedia.org/wiki/Logarithmic_integral_function#Asymptotic_expansion