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

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The discussion centers on proving the relationship between the functions g(x) and f(x) defined in terms of prime numbers congruent to 3 modulo 10. It establishes that f(40) equals 39 and g(40) approximates 6.8, derived from the logarithmic summation of specific primes. The proof utilizes Abel's summation formula to show that g(x) can be expressed as f(x)log(x) minus an integral term. Additionally, there is a discussion on asymptotic behavior, leading to the conclusion that g(x) is asymptotically equivalent to (1/4)x as x approaches infinity. The conversation highlights the importance of precise definitions and calculations in number theory.
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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*}
 
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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*}
 
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.##
 
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
$$
\lim_{x \to \infty}\dfrac{g(x)}{x}=\lim_{x \to \infty}\left(\dfrac{1+(1/x)}{4}-\dfrac{1}{4}\cdot \dfrac{\pi(x)}{x}\right)=\dfrac{1}{4}-\lim_{x \to \infty}\dfrac{1}{4\log(x)}=\dfrac{1}{4}
$$
 
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))) ##?
 
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