The theorem on the Euler Maclaurin Summation formula usually involves the term $x-\lfloor x\rfloor-\frac{1}{2}$, however in some other text, I sometimes also see the term $x-\lfloor x\rfloor+1$ instead.
I came across this issue after looking at two different derivation of the formula /The first proof came from the book: An introduction to Sequences, Series, and Improper Integrals by: O.E. Stanaitis pp.74-75
If $f(x)$ possesses a continuous derivative $f'(x)$ on $[1,n]$, then
$\sum_{s=1}^{n} f(x) = \int_{1}^{n} f(x)dx + \int_{1}^{n} (x-\lfloor x\rfloor- \frac{1}{2})f'(x)dx + \frac{1}{2} [f(1)+f(n)]$
The proof goes as follows:
The obvious equality
$\int_{\mu}^{\mu +1} (x-\mu- \frac{1}{2})f'(x)dx=(x-\mu- \frac{1}{2})f(x)\Big|_{x=\mu}^{x=\mu+1} - \int_{\mu}^{\mu +1} f(x)dx$ obtained by partial integration holds for $\mu=1,2,\dots,n-1.$
In the integrand on the left we can put $\mu=\lfloor x\rfloor$ for $\mu \leq x < \mu + 1.$ Since $x-\lfloor x\rfloor-\frac{1}{2}$ differs from $x-\mu-\frac{1}{2}$ only at one point $x=\mu +1$ on the intervale $[\mu, \mu+1],$ and this does not change the value of the integral, we get
$\frac{1}{2} [f(\mu)+f(\mu+1)]=\int_{\mu}^{\mu + 1} f(x)dx + \int_{\mu}^{\mu + 1} (x-\lfloor x\rfloor - \frac{1}{2})f'(x)dx$
Addition of the relations for $\mu=1,2,\dots,n-1,$ and then of the term $\frac{1}{2} [f(1)+f(n)]$ to both sides yields the Euler's summation formula.
In the book Divergent Series by G.H. Hardy, pp.318-319, "Euler Maclaurin summation" in search box, He present the formula as follows. I will reproduce the first portion of the proof since my question pertains to that.
We suppose first that $0 < a \leq 1,$ that $f'(x)$ is continuous for $x \geq a,$ that $f > 0,$ $f' <0,$ and that $f\rightarrow 0$ when $x \rightarrow \infty.$ Then $\int_{1}^{x} |f'(t)|dt=-\int_{1}^{x} f'(t)dt=f(1)-f(x)\rightarrow f(1)$ when $x\rightarrow \infty,$ so that $\int_{1}^{\infty} |f'(t)|dt=f(1) <\infty.$
if $y=x-[x],$ so that $y=x-m+1$ for $m-1 \leq x < m,$ then $0 \leq y < 1$ and $J= \int_{1}^{\infty} yf'(x)dx$ is absolutely convergent. Also
$j_m = f(m)-\int_{m-1}^{m} f(x)dx=\int_{m-1}^{m} \{f(m)-f(x)\}dx=\int_{m-1}^{m} \{f(m)-f(x)\}\frac{dy}{dx}dx= \int_{m-1}^{m} yf'(x)dx,$
$\sum_{m=2}^{n} f(m)-\int_{1}^{n} f(x)dx=\sum_{m=2}^{n} j_m = \int_{1}^{n} (x-[x])f'(x)dx.$ It follows that, if $F(x)=\int_{a}^{x} f(t)dt,$ $(*)$ then $\sum_{m=1}^{n} f(m)-F(n) \rightarrow f(1)-F(1) + J$ $(**)$ when $n\rightarrow \infty$ If our conditions are satisfied for every $a > 0,$ and $f(x)$ is integrable down to $0,$ we may take $a=0$ in $(*)$ and $(**).$
{Hardy uses $m$ to denote $\lfloor x \rfloor$, since if $m=\lfloor x \rfloor$. $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1,$ then $\lfloor x \rfloor -1 \leq x < \lfloor x \rfloor,$}
What I would like clarification on is this; does it matter that for the greatest integer function $x-\lfloor x\rfloor$, whether the fraction $\frac{1}{2}$ is subtracted from $x-\lfloor x\rfloor$ or a value $\frac{1}{2} \leq \theta < 1$ is added to $x-\lfloor x\rfloor$.
Thank you in advance.
