Question about Euler's summation formula as used in Apostol ANT Given Euler's summation formula in Apostol ANT Theorem 3.1 $$\sum_{y \lt n \leq x} f (n) = \int_y^x f (t) dt + \int_y^x(t- [t])f'(t)dt +f(x)([x]-x) - f(y)([y]-y)$$ Apostol calculates $\sum_{n \leq x} \frac {1}{n} $ in Theorem 3.2a as follows: $$\sum_{n \leq x} \frac {1}{n} = \int_1^x \frac{dt}{t} - \int_1^x \frac{t-[t]}{t^2}dt + 1 - \frac{x-[x]}{x}$$ ( and continues from there ).
My question is about the term $1$. Please explain how this term was calculated. 
 A: Apostol gives for $f(t) = \frac 1t$, $y = 1$, that$\def\fl#1{\left\lfloor#1\right\rfloor}$
$$ \sum_{1 < n \le x} \frac 1n = \int_1^x \frac{dt}t + \int_1^x (t - \fl t)f'(t) \,dt
  + \frac 1x \cdot (\fl x - x) - \frac 11(\fl 1 - 1)$$
Now $f'(t) = -\frac 1{t^2}$, hence 
$$ \sum_{1 < n \le x} \frac 1n = \int_1^x \frac{dt}t - \int_1^x \frac{t - \fl t}{t^2}dt - \frac{x - \fl x}x $$
Now add 1 to both sides.
A: On the left hand side you have $\sum_{y<n\le x}f(n)$. So when using the formula to calculate $\sum_{n\le x}\frac1n$ we must rewrite as $1+\sum_{2<n\le x}\frac1n$.
A: Before I posted this question I had never seen Apostol ' s book on advanced calculus 'Mathematical Analysis' (MA). I have now and it seems to me that difficulties with ANT can be handled by MA. For example I did not know that Euler summation was an application of the Riemann-Stieltjes integral. In MA the Riemann-Stieltjes integral is explained as well as Euler summation as an application. Studying that first will make the first paragraphs of Chater 3 of ANT much more accessible. - See http://www.goodreads.com/book/show/53800.Mathematical_Analysis 
A: I think it's simply a matter of taking $y = 1$ and adding $f(1)$ to both sides of Euler's identity, taking $f(t) = 1/t$ as Apostol says.
