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Given Abel's Identity according to Apostol [1] it follows that,

$\sum_{y <n\leq y} a(n) f(n) = A(x) f(x) - A(y) f(y) - \int\limits_y^{x} A(t) f'(t) dt\quad\quad\quad$

where $A(x) = \sum_{n\leq y} a(n)$

I am having trouble understanding some steps in the proof namely line 4, 5 & 6 below that concerns the Riemann sum to integral that he performs; I cannot understand how he transforms the sum to integral shown in $\color{blue} {blue}$ in the following proof:

PROOF. Let $k=[x]$ and Let $m=[y]$ so that $A(x)=A(k)$ and $A(y)=A(m).$

Then

$$\sum_{y <n\leq x} a(n) f(n) = \sum_{n=m+1}^{k} a(n) f(n) = \sum_{n=m+1}^{k} [A(n)-A(n-1)] f(n)$$

$$= \sum_{n=m+1}^{k} A(n) f(n)-\sum_{n=m}^{k-1} A(n) f(n+1)$$

$$= \sum_{n=m+1}^{k-1} A(n) [f(n)-f(n+1)] + A(k) f(k) - A(m) f(m+1)$$

$$= -\sum_{n=m+1}^{k-1} A(n) \color{blue}{\int_n^{n+1} f(t)'dt} + A(k) f(k) - A(m) f(m+1)$$

$$= -\sum_{n=m+1}^{k-1} \color{blue}{\int_n^{n+1} A(t)f(t)'dt} + A(k) f(k) - A(m) f(m+1)$$

$$= - \color{blue}{\int_{m+1}^{k} A(t)f(t)'dt} + A(x) f(x) - \color{blue}{\int_{k}^{x} A(t)f(t)'dt} - A(y)f(y) - \color{blue}{\int_{y}^{m+1} A(t)f(t)'dt}$$

$$= A(x) f(x) - A(y) f(y) - \int\limits_y^{x} A(t) f'(t) dt$$

[1] Apostol page 77 "Tom M. Apostol-Introduction to Analytic Number Theory-Springer (1976)

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There is a minor mistake in your definition of $A(x)$, it should be

$$A(x) = \sum_{n \le \color{red}x} a(n)$$

From the third line to the fourth line, it is due to

$$\int_n^{n+1}f(t)' \,dt = f(n+1)-f(n) $$

From the fourth line to the fifth line, it is due to

$$\forall t \in [n, n+1), A(t) = A(n) $$

Hence

$$\int_n^{n+1} A(t) f(t)' \, dt= \int_n^{n+1} A(n) f(t)' \, dt = A(n)\int_n^{n+1} f(t)' \, dt.$$

From the fifth line to the sixth line, the first term is just

$$\sum_{n=m+1}^{k-1} \int_n^{n+1} A(t)f(t)' \, dt = \int_{m+1}^k A(t) f(t)' \, dt$$

We are just splitting the integral up.

The second and third term is due to

$$\forall t \in [k,x], A(t)=A(k)=A(x),$$

Hence

$$\int_k^x A(t) f(t)' \, dt = A(k) \int_k^x f(t)' dt = A(k) [f(x)-f(k)]=A(k)f(x)-A(k)f(k)=A(x)f(x)-A(k)f(k).$$

For the fourth and fifth term, it is due to

$$\forall t \in [y, m+1), A(t) = A(y)=A(m),$$

Hence,

$$\int_y^{m+1} A(t)f(t)' \, dt = A(y) \int_y^{m+1 } f(t)' \, dt = A(y) [f(m+1)-f(y)]=A(m)f(m+1)-A(y)f(y)$$

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  • $\begingroup$ thanks for the detailed explaination but I wonder why $A(k)f(x)-A(k)f(k)=A(x)f(x)-A(k)f(k)$? Also isn't there a A(x) missing outside the LHS of $\sum_{n=m+1}^{k-1} \int_n^{n+1} f(t)' \, dt = \int_{m+1}^k A(t) f(t)' \, dt$? $\endgroup$ – onepound Oct 4 '18 at 8:11
  • $\begingroup$ Hi, thanks for pointing the error. we have $A(k)=A(x)$ since $k = \lfloor x \rfloor$ $\endgroup$ – Siong Thye Goh Oct 4 '18 at 8:36
  • $\begingroup$ okay but would that not make $A(k)f(x)−A(k)f(k)$ equal zero if $A(k)=A(x) $? $\endgroup$ – onepound Oct 4 '18 at 17:05
  • $\begingroup$ $f(x)$ need not be equal to $f(k)$. so it need not be zero. $\endgroup$ – Siong Thye Goh Oct 4 '18 at 17:26

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