# Why is $\sum a_nf(n) = \int_0^xf(t)~d(A(t))$?

This equation is a part of Abel's summation formula

$$a_n$$ is a sequence, $$f$$ is a real differentiable function such that $$f'$$ is Riemann integrable.$$A(x) = \sum_{1\le n\le x}a_n$$

I don't see why is $$\sum_{1\le n\le x} a_nf(n) = \int_0^xf(t)~d(A(t))$$

Isn't $$d(A(t)) = 0$$, because $$A(t+h)-A(t) = 0,\text{for t \to 0}$$

$$A$$ is a right-continuous step-function. It is constant on the intervals $$[n-1, n)$$ for all non-negative integers $$n$$, with a “jump height” equal to $$a_n$$ at $$t=n$$.
Now consider partitions $$P = \{ 0 = t_0 < t_1 < \ldots < t_N = x \}$$ with $$\operatorname{norm}(P) = \max(t_{j} - t_{j-1}) < 1$$, and approximation sums $$S(P, f, \alpha) = \sum_{j=1}^N f(c_j)\,(\alpha(t_j) - \alpha(t_{j-1}))$$ with $$c_j \in [t_{j-1}, t_j]$$.
We have $$\alpha(t_j) - \alpha(t_{j-1}) = a_n$$ if the interval $$(t_{j-1}, t_j]$$ contains an integer $$n$$, and $$\alpha(t_j) - \alpha(t_{j-1}) = 0$$ otherwise. It follows that $$S(P, f, \alpha) = \sum_{n \le x} f(c_{j_n}) \, a_n$$ where $$j_n$$ is the index of the interval $$(t_{j-1}, t_j]$$ containing the integer $$n$$.
Now if the norm of the partition approaches zero then $$c_{j_n} \to n$$ and therefore $$f(c_{j_n}) \to f(n)$$ for all the (finitely many) $$n$$ in the interval $$[0, x]$$. This shows that $$\int_0^x f(t) \, d(A(t)) = \sum_{1\le n\le x} f(n) \, a_n \, .$$
Remark: Only the continuity of $$f$$ is needed for this conclusion, more precisely only the continuity of $$f$$ at the integers in the range $$[0, x]$$. The differentiability requirement comes into play when integration by parts is applied to $$\int_0^x f(t) \, d(A(t))$$, which then leads to the Abel's formula.