Th: Consider a sequence ${ (a _n) \subseteq \mathbb{R} }.$
Let ${ \mathbb{R} _{\gt 0} \overset{f}{\to} \mathbb{R} }$ be a ${ \mathcal{C} ^{1} }$ function, and ${ (x _n) \subseteq \mathbb{R} _{\gt 0} }$ a seq with ${ x _n \nearrow \infty }.$ Using ${ (x _n)}$ for indexing gives functions ${ S(X) := \sum _{x _n \leq X} a _n }$ and ${ S _f (X) := \sum _{x _n \leq X} f(x _n) a _n }$ (for every ${ X },$ there are only finitely many ${ n }$ with ${ x _n \leq X },$ making these sums finite).
Now ${ S _f (X) = S(X) f(X) - \int _{x _1} ^{X} S(t) f’(t) dt }$ for ${ X \gt 0 .}$
Pf: When ${ 0 \lt X \lt x _1 }$ both sides are ${ 0 },$ so let ${ x _1 \leq X }.$ We have ${ S _f (X) }$ ${ = \sum _{x _n \leq X} f(x _n) a _n }$ ${ = \sum _{x _n \leq X} \left( f(X) - \int _{x _n} ^{X} f’(t) dt \right) a _n }$ ${ = S(X) f(X) - {\color{green}{\sum _{x _n \leq X} \int _{x _n} ^{X} a _n f’(t) dt} } .}$
Defining ${ g _n (t) }$ to be ${ a _n f ‘(t) }$ when ${ x _n \leq t }$ and ${ 0 }$ when ${ - \infty \lt t \lt x _n },$ we see ${ {\color{green}{\sum _{x _n \leq X} \int _{x _n} ^{X} a _n f’(t) dt} } }$ ${ = \sum _{x _n \leq X} \int _{x _1} ^{X} g _n (t) dt }$ ${ = \int _{x _1} ^{X} \left( \sum _{x _n \leq X} g _n (t) \right)dt }$ ${ = \int _{x _1} ^{X} \left( \sum _{x _n \leq X} \mathbb{I}[ x _n \leq t ] a _n f’(t) \right) dt }$ ${ = \int _{x _1} ^{X} \left( \sum _{x _n \leq X} \mathbb{I}[x _n \leq t] a _n \right) f’(t) dt }$ ${ = \int _{x _1} ^{X} \left( \sum _{x _n \leq t} a _n \right) f ‘(t) dt }$ ${ = \int _{x _1} ^{X} S(t) f’(t) dt },$ as needed.

${ x _n = n }$ case is commonly used.

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EDIT (20/12)

(?) Looks like one could replace ${ x _n \nearrow \infty }$ with ${ x _n \to \infty }$

Repetition

Consider a sequence ${ (a _n) \subseteq \mathbb{R} }.$
Let ${ \mathbb{R} _{\gt 0} \overset{f}{\to} \mathbb{R} }$ be a ${ \mathcal{C} ^{1} }$ function, and ${ (x _n) \subseteq \mathbb{R} _{\gt 0} }$ a seq with ${ x _n \to \infty }.$ Using ${ (x _n)}$ for indexing gives functions ${ S(X) := \sum _{x _n \leq X} a _n }$ and ${ S _f (X) := \sum _{x _n \leq X} f(x _n) a _n }$ (for every ${ X },$ there are only finitely many ${ n }$ with ${ x _n \leq X },$ making these sums finite).
Also ${ x _{\ell} := \min _{n \geq 1} x _n }$ exists (Pick ${ N }$ such that ${ x _n \gt x _1 }$ for all ${ n \geq N }.$ Now ${ x _{\ell} = \min\lbrace x _1, \ldots, x _N \rbrace }$ satisfies ${ x _{\ell} \leq x _n }$ for all ${ n }$).

Th: ${ S _f (X) = S(X) f(X) - \int _{x _{\ell} } ^{X} S(t) f’(t) dt }$ for ${ X \gt 0 }.$
Pf: When ${ 0 \lt X \lt x _{\ell} }$ both sides are ${ 0 ,}$ so let ${ x _{\ell} \leq X }.$ We have ${ S _f (X) }$ ${ = \sum _{x _n \leq X} f(x _n) a _n }$ ${ = \sum _{x _n \leq X} \left( f(X) - \int _{x _n} ^{X} f’(t) dt \right) a _n }$ ${ = S(X) f(X) - {\color{green}{\sum _{x _n \leq X} \int _{x _n} ^{X} a _n f’(t) dt} } .}$
Defining ${ g _n (t) }$ to be ${ a _n f ‘(t) }$ when ${ x _n \leq t }$ and ${ 0 }$ when ${ -\infty \lt t \lt x _n },$ we get ${ {\color{green}{\sum _{x _n \leq X} \int _{x _n} ^{X} a _n f’(t) dt} } }$ ${ = \sum _{x _n \leq X} \int _{x _{\ell} } ^{X} g _n (t) dt }$ ${ = \int _{x _{\ell} } ^{X} \left( \sum _{x _n \leq X} g _n (t) \right)dt }$ ${ = \int _{x _{\ell} } ^{X} \left( \sum _{x _n \leq X} \mathbb{I}[ x _n \leq t ] a _n f’(t) \right) dt }$ ${ = \int _{x _{\ell} } ^{X} \left( \sum _{x _n \leq X} \mathbb{I}[x _n \leq t] a _n \right) f’(t) dt }$ ${ = \int _{x _{\ell} } ^{X} \left( \sum _{x _n \leq t} a _n \right) f ‘(t) dt }$ ${ = \int _{x _{\ell} } ^{X} S(t) f’(t) dt },$ as needed.