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

• Math100
In summary, we used Abel's summation formula to find the values of the functions ##f(40)## and ##g(40)##, which are equal to 39 and approximately 6.8, respectively. We also proved the equation ##g(x)=f(x)\log {x}-\int_{2}^{x}t^{-1}f(t)dt## using this formula.

#### 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*}

Math100 said:
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.

Math100 said:
a) ## f(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}1=3+13+23=39 ##

Math100 said:
## 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##
Math100 said:
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
fresh_42 said:
##\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 ##.

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.##

Math100 said:
How did you get ## g(40)\approx 6.8 ##? I thought it's ## g(40)\approx 2.95 ##.
Math100 said:
## 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
fresh_42 said:
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 ##?

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
fresh_42 said:
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))) ##?

Math100