I am reading "An Elementary View of Euler's Summation Formula" by Tom M. Apostol.
Let $f$ be a function with a countinuous derivative on the interval $[1,n]$.
We consider $\sum_{k=1}^{n-1}f(k)-\int_{1}^{n}f(x)\,dx$. (See the picture below. $f(x)=\frac{1}{x}\sin(\frac{x}{5})$ and $n=50$.)
$$\sum_{k=1}^{n-1}f(k)-\int_{1}^{n}f(x)\,dx=\sum_{k=1}^{n-1}\int_{k}^{k+1} f(k)\,dx-\sum_{k=1}^{n-1}\int_{k}^{k+1}f(x)\,dx\\=\sum_{k=1}^{n-1}\int_{k}^{k+1}\{f(k)-f(x)\}\,dx.$$
$$\int_{k}^{k+1}\{f(k)-f(x)\}\,dx=\int_{k}^{k+1} (x-k-1)^{'}\{f(k)-f(x)\}\,dx\\=[(x-k-1)\{f(k)-f(x)\}]_{k}^{k+1}+\int_{k}^{k+1} (x-k-1)f^{'}(x)\,dx\\=\int_{k}^{k+1}(x-k-1)f^{'}(x)\,dx=\int_{k}^{k+1}(x-[x]-1)f^{'}(x)\,dx.$$
So, $$\sum_{k=1}^{n-1} f(k)-\int_{1}^{n} f(x)\,dx=\sum_{k=1}^{n-1}\int_{k}^{k+1}(x-[x]-1)f^{'}(x)\,dx\\=\sum_{k=1}^{n-1}[\int_{k}^{k+1}(x-[x])f^{'}(x)\,dx-\int_{k}^{k+1} f^{'}(x)\,dx]\\=\int_{1}^{n}(x-[x])f^{'}(x)\,dx-\int_{1}^{n}f^{'}(x)\,dx\\=\int_{1}^{n} (x-[x])f^{'}(x)\,dx+f(1)-f(n).$$
So, $$\sum_{k=1}^{n} f(k)=\int_{1}^{n} f(x)\,dx+\int_{1}^{n} (x-[x])f^{'}(x)\,dx+f(1).$$
Since $$\int_{1}^{n} (x-[x])f^{'}(x)\,dx\\=\int_{1}^{n} (x-[x]-\frac{1}{2})f^{'}(x)\,dx+\int_{1}^{n}\frac{1}{2} f^{'}(x)\,dx\\=\int_{1}^{n}(x-[x]-\frac{1}{2})f^{'}(x)\,dx+\frac{1}{2}\{f(n)-f(1)\},$$
$$\sum_{k=1}^{n} f(k)=\int_{1}^{n} f(x)\,dx+\int_{1}^{n} (x-[x]-\frac{1}{2})f^{'}(x)\,dx+\frac{1}{2}\{f(n)+f(1)\}.\tag{8}\label{8}$$
The factor $x-[x]-\frac{1}{2}$ has the value $-\frac{1}{2}$ when $x$ is an integer. We modify this factor slightly to make it vanish at the integers, a property that is desirable when we integrate by parts. To do this we introduce $P_1(x)$, the first Bernoulli function:
$P_1(x)= \begin{cases} x-[x]-\frac{1}{2} & \text{if $x\neq$ integer} \\ 0 & \text{if $x=$ integer.} \end{cases}\tag{9}$ The error integral does not change if the factor $x-[x]-\frac{1}{2}$ is replaced by $P_1(x)$ because the two factors differ only at the integers. Therefore $\eqref{8}$ can be written as $$\sum_{k=1}^{n} f(k)=\int_{1}^{n} f(x)\,dx+\int_{1}^{n} P_1(x)f^{'}(x)\,dx+\frac{1}{2}\{f(n)+f(1)\}.\tag{10}\label{10}$$
Note the cotrast between $\eqref{10}$ and (3), which explicitly displays the generalized Euler's constant $C(f)$. To make $\eqref{10}$ resemble (3) more closely, we assume that the improper integral $\int_{1}^{\infty} P_1(x)f^{'}(x)\,dx$ converges. Then we can write $$\int_{1}^{n}P_1(x)f^{'}(x)\,dx=\int_{1}^{\infty}P_1(x)f^{'}(x)\,dx-\int_{n}^{\infty}P_1(x)f^{'}(x)\,dx,$$ and $\eqref{10}$ takes the form $$\sum_{k=1}^{n}f(k)=\int_{1}^{n}f(x)\,dx+C(f)+E_f(n),\tag{11}\label{11}$$ where $$C(f)=\frac{1}{2}f(1)+\int_{1}^{\infty}P_1(x)f^{'}(x)\,dx\tag{12}\label{12}$$ and $$E_f(n)=\frac{1}{2}f(n)-\int_{n}^{\infty} P_1(x)f^{'}(x)\,dx.$$
Eq. $\eqref{11}$ has exactly the same form as (3), but $\eqref{11}$ is more general because $f$ is not required to be positive or monotonic. The only restrictions on $f$ are continuity of $f^{'}$ and convergence of the improper integral $$\int_{1}^{\infty}P_1(x)f^{'}(x)\,dx.\tag{13}\label{13}$$
The improper integral in $\eqref{13}$ converges if, and only if, $$\lim_{n\to\infty}\int_{n}^{\infty}P_1(x)f^{'}(x)\,dx=0.\tag{14}\label{14}$$
A sufficient condition for convergence is that $\int_{1}^{\infty}|f^{'}(x)|\,dx$ converges, or equivalently, that $$\lim_{n\to\infty}\int_{n}^{\infty}|f^{'}(x)|\,dx=0.\tag{15}\label{15}$$ To see this, note that the Bernoulli function $P_1(x)$ is bounded; in fact, Figure 3 shows that $|P_1(x)|\leq\frac{1}{2}$ for all $x$, so $\eqref{14}$ follows from $\eqref{15}$.
Example. When $f(x)=1/x$ we have $f^{'}(x)=-1/x^2$ and $$\int_{n}^{\infty}|f^{'}(x)|\,dx=\int_{n}^{\infty}\frac{1}{x^2}\,dx=\frac{1}{n}.$$
Therefore $\eqref{15}$ is satisfied and $\eqref{12}$ expresses Euler's constant as an integral: $$C=\frac{1}{2}-\int_{1}^{\infty}\frac{P_1(x)}{x^2}\, dx.$$
I cannot understand what the author wants to say.
I think $\eqref{14}$ means $$\int_{n}^{\infty}P_1(x)f^{'}(x)\,dx$$ exists for any $n\in\{1,2,\dots\}$ and $$\lim_{n\to\infty}\int_{n}^{\infty}P_1(x)f^{'}(x)\,dx=0$$ holds.
So, I think we don't use $\eqref{14}$ to show $\eqref{13}$ converges because if $\eqref{14}$ holds, then $\eqref{13}$ converges automatically.
The author wrote $\eqref{14}$ follows from $\eqref{15}$.
I think it is true.
But I think we don't use $\eqref{15}$ to show $\eqref{13}$ converges because if $\eqref{15}$ holds, then $\eqref{14}$ holds.
So, $\eqref{13}$ converges automatically.
I think if we want to show $\eqref{13}$ converges, then we will show $\int_{1}^{\infty}|f^{'}(x)|\,dx$ converges.
What does the author want to say?
Please explain me.