In applications of Euler's summation formula to find the asymptotic behaviour of something like $\sum_{n\leq x}f(n)$, one typically gets integrals of the form $$\int_1^x (t-[t])f'(t)dt$$ where $[t]$ denotes the integer part of $t$.
I've two questions, relating particularly to cases where the function $f(x)$ involves some term in $\log x$. As a simple (and well-known) example, lets take $f(x) = \frac{\log x}{x}$.
The standard method is to split the integral, $\int_1^x = \int_1^\infty - \int_x^\infty$, and approximate each part. This leads to the conclusion that $$\int_1^x(t-[t])f'(t)dt = K + O(f(x))$$ for some constant $K$ (equal to the first integral). However, what's wrong with doing the following? $$\int_1^x (t-[t])f'(t)dt \leq \int_1^x f'(t)dt = f(x)-f(1)$$ If $f(1)\neq 0$, then this tells us that the integral is $O(1)$, so it's a less good approximation for the integral than if we'd split it up. However, when $f(x)=\frac{\log x}x$, we have $f(1)=0$, so the constant term has disappeared and thus we can conclude $$\int_1^x (t-[t])\left(\frac{\log t}t\right)'dt = O\left(\frac{\log x}x\right).$$ Since there is now no constant $K$, there must be something wrong in this argument, but what?
When $f(x)$ involves a logarithm, it's typically stated in the question that the approximation is valid for $x\geq 2$. Obviously if $x<2$ then $\sum_{n\leq x}f(x)=0$, but where does this condition actually come into play in the solution?