How to prove this using Abel's summation formula?

[Math100]

Homework Statement

Prove that $\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\frac{1}{p}=\frac{1}{4}\log\log {x}+A+O(\frac{1}{\log {x}})$, for some constant $A$.

Relevant Equations

Suppose $gcd(h, k)=1$ for $k>0$. Then, there exists a constant $A$ such that for all $x\geq 2$:
$\sum_{\substack{prime p\leq x \\ p\equiv h\pmod {k}}}\frac{1}{p}=\frac{1}{\varphi(k)}\log\log {x}+A+O(\frac{1}{\log {x}})$.

Abel's summation formula:
If $A(x)=\int_{y}^{x}a(t)dt$, then $\int_{y}^{x}a(t)f(t)dt=A(x)f(x)-A(y)f(y)-\int_{y}^{x}A(t)f'(t)dt$.

Before I apply/use the Abel's summation formula, how should I find $f(x)$?

 

[fresh_42]

Abel's summation formula goes
$$
\sum_{x\leq n\leq y}a_n f(n)=A(y)f(y)-A(x)f(x)-\int_x^y A(t)f'(t)\,dt
$$
with $A(x)=\displaystyle{\sum_{0<n\leq x}}a_n$ where $(a_n)$ is a sequence of real or complex numbers, and $f(t)$ any on $[x,y]$ continuously differentiable function. Your choice.

How do you reference the important equation under "relevant equations"?

https://en.wikipedia.org/wiki/Abel's_summation_formula

 

[Math100]

Abel's summation formula goes
$$
\sum_{x\leq n\leq y}a_n f(n)=A(y)f(y)-A(x)f(x)-\int_x^y A(t)f'(t)\,dt
$$
with $A(x)=\displaystyle{\sum_{0<n\leq x}}a_n$ where $(a_n)$ is a sequence of real or complex numbers, and $f(t)$ any on $[x,y]$ continuously differentiable function. Your choice.

How do you reference the important equation under "relevant equations"?

https://en.wikipedia.org/wiki/Abel's_summation_formula

From the textbook and notes. But if $f(t)$ can be any on $[x, y]$, then what would you choose? How to choose this $f(t)$ function wisely?

 

[fresh_42]

Dirichlet's theorem on arithmetic progressions says
$$
\sum_{\stackrel{p\text{ prime }}{p\equiv h\pmod{k}\, , \,(h,k=1)}}\dfrac{1}{p}=\infty
$$
Your theorem says
$$
S(k,x):=\sum_{\stackrel{p\text{ prime }\leq x}{p\equiv h\pmod{k}\, , \,(h,k=1)}}\dfrac{1}{p}=\dfrac{1}{\varphi(k)}\log\log {x}+A+O\left(\dfrac{1}{\log {x}}\right)
$$
so the boundary $\leq x$ changes everything.

Before we chose $f(t)$ let's look at the sequence, just to get an impression
\begin{matrix}
3& 13& 23& 43& 53& 73& 83& 103& 113& 163& 173& 193& 223& 233\\
263& 283& 293& 313& 353& 373& 383& 433& 443& 463&503& 523& 563& 593\\
613&643 &653& 673& 683& 733& 743& 773& 823& 853& 863& 883& 953& 983\\
1013& 1033& 1063& 1093& 1103& 1123& 1153& 1163& 1193& 1213& 1223& 1283& 1303& 1373\\
1423& 1433& 1453&1483& 1493& 1523& 1543& 1553&\ldots &&&&&&
\end{matrix}

The prime number theorem gives us $\displaystyle{\lim_{x \to \infty}\dfrac{\pi(x)\log(x)}{x}=1}$ where the prime-counting function is given as $\displaystyle{\pi(x)=\sum_{\stackrel{p\text{ prime }}{p\leq x}}}.$ The $n-$th prime number can be approximated by $p(n)\sim n\log(n).$ The error approaches zero for growing $n.$

We set $h=3$ and $k=10.$ This gives us from your theorem
$$
S(10,x)= \dfrac{1}{4}\log\log {x}+A+O\left(\dfrac{1}{\log {x}}\right)
$$
since $\varphi (10)=|\{x\pmod{10}\,|\,(x,10)=1\}|=|\{1,3,7,9\}|=4.$

Do you want to prove the theorem?

 

[Math100]

Dirichlet's theorem on arithmetic progressions says
$$
\sum_{\stackrel{p\text{ prime }}{p\equiv h\pmod{k}\, , \,(h,k=1)}}\dfrac{1}{p}=\infty
$$
Your theorem says
$$
S(k,x):=\sum_{\stackrel{p\text{ prime }\leq x}{p\equiv h\pmod{k}\, , \,(h,k=1)}}\dfrac{1}{p}=\dfrac{1}{\varphi(k)}\log\log {x}+A+O\left(\dfrac{1}{\log {x}}\right)
$$
so the boundary $\leq x$ changes everything.

Before we chose $f(t)$ let's look at the sequence, just to get an impression
\begin{matrix}
3& 13& 23& 43& 53& 73& 83& 103& 113& 163& 173& 193& 223& 233\\
263& 283& 293& 313& 353& 373& 383& 433& 443& 463&503& 523& 563& 593\\
613&643 &653& 673& 683& 733& 743& 773& 823& 853& 863& 883& 953& 983\\
1013& 1033& 1063& 1093& 1103& 1123& 1153& 1163& 1193& 1213& 1223& 1283& 1303& 1373\\
1423& 1433& 1453&1483& 1493& 1523& 1543& 1553&\ldots &&&&&&
\end{matrix}

The prime number theorem gives us $\displaystyle{\lim_{x \to \infty}\dfrac{\pi(x)\log(x)}{x}=1}$ where the prime-counting function is given as $\displaystyle{\pi(x)=\sum_{\stackrel{p\text{ prime }}{p\leq x}}}.$ The $n-$th prime number can be approximated by $p(n)\sim n\log(n).$ The error approaches zero for growing $n.$

We set $h=3$ and $k=10.$ This gives us from your theorem
$$
S(10,x)= \dfrac{1}{4}\log\log {x}+A+O\left(\dfrac{1}{\log {x}}\right)
$$
since $\varphi (10)=|\{x\pmod{10}\,|\,(x,10)=1\}|=|\{1,3,7,9\}|=4.$

Do you want to prove the theorem?

Before I try to prove the theorem, I want to know, how did you get those numbers from $3$ to $1553$ from the sequence?

 

[fresh_42]

Before I try to prove the theorem, I want to know, how did you get those numbers from $3$ to $1553$ from the sequence?

That was the easy part.

I copied the list up to ~1600 from https://de.wikibooks.org/wiki/Primzahlen:_Tabelle_der_Primzahlen_(2_-_100.000),
put it into an editor (textpad), changed the comma (,) globally to ampersand (&), dropped it in a begin-end-matrix environment here, deleted all numbers that did not end on 3, and finally inserted a newline command (\\) every fourteen primes.

If you want to prove the theorem, then why for $(h,k)=(3,10)$ and not generally for $(h,k)$? And wasn't it proven in the book? You said it is from there.

I haven't found the exact version of Dirichlet's prime number theorem that your book lists but the one I did have found is heavy machinery and I seriously doubt that your version or the specification $(h,k)=(3,10)$ makes it a lot easier than that:
https://www.glk.uni-mainz.de/files/2018/08/Nickel.pdf (theorem 4.16, page 23/23)

 

[Math100]

That was the easy part.

I copied the list up to ~1600 from https://de.wikibooks.org/wiki/Primzahlen:_Tabelle_der_Primzahlen_(2_-_100.000),
put it into an editor (textpad), changed the comma (,) globally to ampersand (&), dropped it in a begin-end-matrix environment here, deleted all numbers that did not end on 3, and finally inserted a newline command (\\) every fourteen primes.

If you want to prove the theorem, then why for $(h,k)=(3,10)$ and not generally for $(h,k)$? And wasn't it proven in the book? You said it is from there.

I haven't found the exact version of Dirichlet's prime number theorem that your book lists but the one I did have found is heavy machinery and I seriously doubt that your version or the specification $(h,k)=(3,10)$ makes it a lot easier than that:
https://www.glk.uni-mainz.de/files/2018/08/Nickel.pdf (theorem 4.16, page 23/23)

The following proof below is from the book:

Let $f(x)=\frac{1}{\log {x}}$ and $a(n)=\frac{\log {n}}{n}$ if $n\equiv h\pmod {k}$ is prime
and $0$ otherwise.
By Dirichlet's Theorem, we have $\sum_{n\leq x}a(n)=\frac{1}{\varphi(k)}\log {x}+R(x)$,
where $R(x)=O(1)$.
Applying Abel's summation formula,
$\sum_{\substack{prime p\leq x \\ p\equiv h\pmod {k}}}\frac{1}{p}=\sum_{1<n\leq x}a(n)f(n)$
$=\frac{1}{\log {x}}(\frac{1}{\varphi(k)}\log {x}+O(1))+\int_{2}^{x}\frac{1}{t\log^{2} {t}}(\frac{1}{\varphi(k)}\log {t}+R(t))dt$
$=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}})+\frac{1}{\varphi(k)}\int_{2}^{x}\frac{dt}{t\log {t}}+\int_{2}^{\infty}\frac{R(t)}{t\log^{2} {t}}dt+\int_{x}^{\infty}\frac{R(t)}{t\log^{2} {t}}dt$
$=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}})+\frac{\log\log {x}-\log\log {2}}{\varphi(k)}+C+O(\int_{x}^{\infty}\frac{dt}{t\log^{2} {t}})$
$=\frac{1}{\varphi(k)}\log\log {x}+A+O(\frac{1}{\log {x}})$.

But I don't understand the first part in this proof. Why letting $f(x)=\frac{1}{\log {x}}$ and $a(n)=\frac{\log {n}}{n}$?

 

[fresh_42]

The following proof below is from the book:

Let $f(x)=\frac{1}{\log {x}}$ and $a(n)=\frac{\log {n}}{n}$ if $n\equiv h\pmod {k}$ is prime
and $0$ otherwise.
By Dirichlet's Theorem, we have $\sum_{n\leq x}a(n)=\frac{1}{\varphi(k)}\log {x}+R(x)$,
where $R(x)=O(1)$.

Yes, this was the difficult part that I thought you wanted to prove. It contains the so-called Dirichlet density $\varphi (k)^{-1}$ which isn't easy to see where it comes from.

Applying Abel's summation formula, $\sum_{\substack{prime p\leq x \\ p\equiv h\pmod {k}}}\frac{1}{p}=\sum_{1<n\leq x}a(n)f(n)$
$=\frac{1}{\log {x}}(\frac{1}{\varphi(k)}\log {x}+O(1))+\int_{2}^{x}\frac{1}{t\log^{2} {t}}(\frac{1}{\varphi(k)}\log {t}+R(t))dt$
$=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}})+\frac{1}{\varphi(k)}\int_{2}^{x}\frac{dt}{t\log {t}}+\int_{2}^{\infty}\frac{R(t)}{t\log^{2} {t}}dt+\int_{x}^{\infty}\frac{R(t)}{t\log^{2} {t}}dt$
$=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}})+\frac{\log\log {x}-\log\log {2}}{\varphi(k)}+C+O(\int_{x}^{\infty}\frac{dt}{t\log^{2} {t}})$
$=\frac{1}{\varphi(k)}\log\log {x}+A+O(\frac{1}{\log {x}})$.

But I don't understand the first part in this proof. Why letting $f(x)=\frac{1}{\log {x}}$ and $a(n)=\frac{\log {n}}{n}$?

We want to apply Dirichlet's theorem in its classical form and improve the boundary of its estimation. The classical version reads (with $a_n$ as in your textbook)
$$
\sum_{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{1}{p}}=
\sum_{0<n\leq x} a_n
={\frac{\log(x)}{\varphi (k)}}+R(x)
$$
The problem is that we cannot really work with that sum on the left that contains so many gaps compared to all integers $\{1,2,\ldots,[x]\}.$ So we will start with a replacement
$$
\sum _{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod {k}}}{\frac {1}{p}}=\sum_{n=1}^{[x]}b_n
$$
We would like to set $b_n=1/n$ but we have to consider the gaps. Hence we define
$$
b_n=\begin{cases}1/n&\text{ if }n\leq x \text{ is prime and }n\equiv h{\pmod {k}}\\0&\text{ else }\end{cases}
$$
Our next difficulty is that we only have an estimation
$$
\sum_{0<n\leq x}a_n=\sum _{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{\log p}{p}}
$$
So we have $b_p=\dfrac{1}{p}=\dfrac{a_p}{\log p}$ which also holds for the zeros, so $b_n=\dfrac{1}{n}=\dfrac{a_n}{\log n}$ bridging the gaps. The difference is thus a factor $f(n)=\frac{1}{\log n}$ that we can use in Abel's summation formula since we can choose $f(n)$ as we like as long as it is continuously differentiable. All in all we have now - and note that $a_1=0$
\begin{align*}
\sum_{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{1}{p}}&=
\sum_{n=1}^{[x]}b_n=\sum_{n=1}^{[x]}\dfrac{a_n}{\log n}=\sum_{n=1}^{[x]}a_nf(n)\\
&\stackrel{\text{Abel}}{=}\left(\sum_{n=1}^xa_n\right)f(x)-\left(\sum_{n=1}^1 a_n\right)f(1)-\int_1^x \left(\sum_{n=1}^t a_n\right)f'(t)\,dt \\
&\stackrel{\text{Dirichlet}}{=}\left({\frac{\log(x)}{\varphi (k)}}+R(x)\right)\cdot \dfrac{1}{\log x}-\int_2^x\left({\frac{\log(t)}{\varphi (k)}}+R(t)\right)\left(\dfrac{1}{\log t}\right)'\,dt\\
&= \dfrac{1}{\log x}\left({\frac{\log(x)}{\varphi (k)}}+O(1)\right)+\int_2^x \left(\dfrac{1}{t \log^2 t}\right) \left({\frac{\log(t)}{\varphi (k)}}+R(t)\right)\,dt\\
&\phantom{=}\ldots
\end{align*}

 

[Math100]

Yes, this was the difficult part that I thought you wanted to prove. It contains the so-called Dirichlet density $\varphi (k)^{-1}$ which isn't easy to see where it comes from.We want to apply Dirichlet's theorem in its classical form and improve the boundary of its estimation. The classical version reads (with $a_n$ as in your textbook)
$$
\sum_{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{1}{p}}=
\sum_{0<n\leq x} a_n
={\frac{\log(x)}{\varphi (k)}}+R(x)
$$
The problem is that we cannot really work with that sum on the left that contains so many gaps compared to all integers $\{1,2,\ldots,[x]\}.$ So we will start with a replacement
$$
\sum _{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod {k}}}{\frac {1}{p}}=\sum_{n=1}^{[x]}b_n
$$
We would like to set $b_n=1/n$ but we have to consider the gaps. Hence we define
$$
b_n=\begin{cases}1/n&\text{ if }n\leq x \text{ is prime and }n\equiv h{\pmod {k}}\\0&\text{ else }\end{cases}
$$
Our next difficulty is that we only have an estimation
$$
\sum_{0<n\leq x}a_n=\sum _{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{\log p}{p}}
$$
So we have $b_p=\dfrac{1}{p}=\dfrac{a_p}{\log p}$ which also holds for the zeros, so $b_n=\dfrac{1}{n}=\dfrac{a_n}{\log n}$ bridging the gaps. The difference is thus a factor $f(n)=\frac{1}{\log n}$ that we can use in Abel's summation formula since we can choose $f(n)$ as we like as long as it is continuously differentiable. All in all we have now - and note that $a_1=0$
\begin{align*}
\sum_{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{1}{p}}&=
\sum_{n=1}^{[x]}b_n=\sum_{n=1}^{[x]}\dfrac{a_n}{\log n}=\sum_{n=1}^{[x]}a_nf(n)\\
&\stackrel{\text{Abel}}{=}\left(\sum_{n=1}^xa_n\right)f(x)-\left(\sum_{n=1}^1 a_n\right)f(1)-\int_1^x \left(\sum_{n=1}^t a_n\right)f'(t)\,dt \\
&\stackrel{\text{Dirichlet}}{=}\left({\frac{\log(x)}{\varphi (k)}}+R(x)\right)\cdot \dfrac{1}{\log x}-\int_2^x\left({\frac{\log(t)}{\varphi (k)}}+R(t)\right)\left(\dfrac{1}{\log t}\right)'\,dt\\
&= \dfrac{1}{\log x}\left({\frac{\log(x)}{\varphi (k)}}+O(1)\right)+\int_2^x \left(\dfrac{1}{t \log^2 t}\right) \left({\frac{\log(t)}{\varphi (k)}}+R(t)\right)\,dt\\
&\phantom{=}\ldots
\end{align*}

Just to confirm from where you left off,
\begin{align*}
&\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\int_{2}^{x}(\frac{1}{t\log {t}}\cdot \frac{1}{\varphi(k)}+\frac{R(t)}{t\log^2 {t}})dt\\
&=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\frac{1}{\varphi(k)}\int_{2}^{x}\frac{dt}{t\log {t}}+\int_{2}^{x}\frac{R(t)}{t\log^2 {t}}dt+\int_{x}^{\infty}\frac{R(t)}{t\log^2 {t}}dt\\
&=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\frac{\log\log {x}-\log\log {2}}{\varphi(k)}+C+O(\frac{1}{\log {2}})+O(\frac{1}{\log {x}})\\
&=\frac{1}{\varphi(k)}(1+\log\log {x}-\log\log {2})+C+O(\frac{1}{\log {x}})\\
&=\frac{1}{\varphi(k)}\log\log {x}+\frac{1}{\varphi(k)}(1-\log\log {2})+C+O(\frac{1}{\log {x}})\\
&=\frac{1}{\varphi(k)}\log\log {x}+A+O(\frac{1}{\log {x}})\\
\end{align*}
where $A=\frac{1}{\varphi(k)}(1-\log\log {2})+C$.
Is this correct?

 

[fresh_42]

Just to confirm from where you left off,
\begin{align*}
&\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\int_{2}^{x}(\frac{1}{t\log {t}}\cdot \frac{1}{\varphi(k)}+\frac{R(t)}{t\log^2 {t}})dt\\
&=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\frac{1}{\varphi(k)}\int_{2}^{x}\frac{dt}{t\log {t}}+\int_{2}^{x}\frac{R(t)}{t\log^2 {t}}dt+\int_{x}^{\infty}\frac{R(t)}{t\log^2 {t}}dt\\
&=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\frac{\log\log {x}-\log\log {2}}{\varphi(k)}+C+O(\frac{1}{\log {2}})+O(\frac{1}{\log {x}})\\
&=\frac{1}{\varphi(k)}(1+\log\log {x}-\log\log {2})+C+O(\frac{1}{\log {x}})\\
&=\frac{1}{\varphi(k)}\log\log {x}+\frac{1}{\varphi(k)}(1-\log\log {2})+C+O(\frac{1}{\log {x}})\\
&=\frac{1}{\varphi(k)}\log\log {x}+A+O(\frac{1}{\log {x}})\\
\end{align*}
where $A=\frac{1}{\varphi(k)}(1-\log\log {2})+C$.
Is this correct?

Looks ok, modulo a few typos. Let me explain the calculation (explanations in brackets $[\,\cdot\,]$ at the end of a line):

\begin{align*}
\sum_{{p\ {\text{ prime }} \atop p\leq x} \atop p\equiv h{\pmod{k}}}{\frac{1}{p}}&=
\sum_{n=1}^{[x]}b_n=\sum_{n=1}^{[x]}\dfrac{a_n}{\log n}=\sum_{n=1}^{[x]}a_nf(n)\\
&\stackrel{\text{Abel}}{=}\left(\sum_{n=1}^xa_n\right)f(x)-\left(\sum_{n=1}^1 a_n\right)f(1)-\int_1^x \left(\sum_{n=1}^t a_n\right)f'(t)\,dt \\
&\stackrel{\text{Dirichlet}}{=}\left({\frac{\log(x)}{\varphi (k)}}+R(x)\right)\cdot \dfrac{1}{\log x}-\int_2^x\left({\frac{\log(t)}{\varphi (k)}}+R(t)\right)\left(\dfrac{1}{\log t}\right)'\,dt\\
&= \dfrac{1}{\log x}\left({\frac{\log(x)}{\varphi (k)}}+O(1)\right)+\int_2^x \left(\dfrac{1}{t \log^2 t}\right) \left({\frac{\log(t)}{\varphi (k)}}+R(t)\right)\,dt\\
&= \dfrac{1}{\varphi (k)}+O\left(\dfrac{1}{\log x}\right)+\ldots \quad \left[\dfrac{1}{\log x}\cdot O(1)=\dfrac{1}{\log x}\cdot C=\dfrac{C}{\log x}=O\left(\dfrac{1}{\log x}\right)\right]\\[12pt]
\ldots &+\dfrac{1}{\varphi (k)}\int_2^x \dfrac{1}{t \log t}\,dt+ C \int_2^x \dfrac{1}{t \log^2 t}\,dt \;\ldots\quad \left[R(t)=O(1)=C\right]\\[12pt]
&=\dfrac{1-\log\log 2}{\varphi (k)}+O\left(\dfrac{1}{\log x}\right)+\dfrac{\log\log x}{\varphi (k)}+\ldots \\[12pt]
&\quad\quad\quad\left[\dfrac{d}{dt}\log (\log (t))=\dfrac{1}{t\log t}\Rightarrow \int_2^x \dfrac{1}{t \log t}\,dt=\log (\log (x))-\log\log (2)\right]\\[12pt]
&\quad\quad\quad\left[\text{ and }\log\log 2 \text{ joins the first term}\right]\\[12pt]
\ldots &+ C\cdot \left(\underbrace{\int_2^\infty \dfrac{1}{t \log^2 t}\,dt}_{=\dfrac{1}{\log 2}}- \underbrace{\int_x^\infty \dfrac{1}{t \log^2 t}\,dt}_{=\dfrac{1}{\log x}}\right)\;\ldots\quad \left[\int_a^b =\int_a^\infty -\int_x^\infty \right]\\[12pt]
&=\dfrac{1-\log\log 2}{\varphi (k)}+O\left(\dfrac{1}{\log x}\right)+\dfrac{\log\log x}{\varphi (k)}+\underbrace{\dfrac{C}{\log 2}}_{=:A}-\dfrac{C}{\log x}\\[12pt]
&=\dfrac{\log\log x}{\varphi (k)}+A+O\left(\dfrac{1}{\log x}\right)+\dfrac{1-\log\log 2}{\varphi (k)}\;\ldots\quad\\[12pt]
&\quad\quad\quad\left[O\left(\log^{-1}(x)\right)-C\log^{-1}(x)=(C'-C)\log^{-1}(x)=O(\log^{-1}(x))\right]
\end{align*}
I hesitate to decide where to put $\dfrac{1-\log\log 2}{\varphi (k)}$ to.

If we pushed it in the first term $\dfrac{\log\log x}{\varphi (k)}$ we would get $\gamma \dfrac{\log\log x}{\varphi (k)}=O\left(\dfrac{\log\log x}{\varphi (k)}\right)$ which would ruin our nice first term, which is the entire reason we did this here.

We cannot put it into $O\left(\dfrac{1}{\log x}\right)$ because this gets small for increasing $x$ whereas $\dfrac{1-\log\log 2}{\varphi (k)}$ is independent of $x$. It would ruin the boundary.

If we put it in $A$, then we cheat a little bit. Why do we have $\varphi (k)$ explicitly in the first term if we treat it as a constant in $A$? $k$ is a constant, so $A$ is probably a good place to be. Thus
$$
A=\dfrac{C}{\log 2}+\dfrac{1-\log\log 2}{\varphi (k)}
$$
$C$ came in as $R(t)=O(1)$ which is already quite arbitrary. I guess that's why it's best that $A$ swallows $\dfrac{1-\log\log 2}{\varphi (k)}$ as another additive constant, est. $+1.37\cdot \varphi (k).$