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Proof from Apostol: (Annotations mine)

If $f$ has a continuous derivative $f'$ on the interval $[y, x]$, where $0 < y < x$, then

$$\sum\limits_{y < n \le x} f(n) = \int_y^x f(t) dt + \int_y^x(t - [t]) f'(t) dt$$ $$ + f(x)([x] - x) - f(y)([y] - y) \ \ \ \ \ \ \ \ \ \ (1)$$

Proof: Let $m = [y], k = [x]$. For integers $n$ and $n-1$ in [y,x] we have

$$\int_{n-1}^n [t]f'(t) dt =\int_{n-1}^n (n-1)f'(t)dt = (n-1)\{f(n) - f(n-1)\} $$ $$ = \{nf(n) - (n-1)f(n-1)\} - f(n) \ \ \ \ \ (2)$$

Summing from $n = m+1$ to $k$, we find:

$$ \int_m^k [t]f'(t) dt = \sum\limits_{n=m+1}^k\{nf(n) - (n-1)f(n-1)\} -\sum\limits_{y<n\le x} f(n) \ \ \ (3)$$

Hence $$\sum\limits_{y<n\le x} f(n) = -\int_m^k [t]f'(t) dt +kf(k) - mf(m)\ \ \ (4)$$ $$= -\int_m^k [t]f'(t) dt +kf(x) - mf(y) \ \ \ (5)$$

Integration by parts gives us:

$$\int_y^x f(t) dt = xf(x) - y f(y) - \int_y^x tf'(t) dt \ \ \ (6) $$

and when this is combined with $(5)$ we get $(1)$. End of Proof.

Questions: I understand most of the proof except for two things:

How could we go from $(4)$ to $(5)$, i.e., replace $f(k)$ by $f(x)$. Is this just an integer to real type conversion?

I don't get the last sentence of the proof, "and when this[referring to $(6)$] is combined with $(5)$ we get $(1)$". How?

Thanks

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1 Answer

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Here are the correct steps leading to the correct form of (5) where the integration limits should be $x$ and $y$.

Starting with (2) which you obtained correctly,

$$\int_{n-1}^n [t]f'(t) dt = \{nf(n) - (n-1)f(n-1)\}- f(n),$$

sum from $n = m+2$ to $k$ to obtain

$$\int_{m+1}^k [t]f'(t) dt = \sum\limits_{n=m+2}^k\{nf(n) - (n-1)f(n-1)\} -\sum_{n=m+2}^k f(n) \\ = kf(k) - (m+1)f(m+1) - \sum_{n=m+2}^k f(n) \\ = kf(k) - mf(m+1) - \sum_{n=m+1}^k f(n).$$

Hence,

$$\sum_{y < n \leqslant x} f(n) = -\int_{m+1}^k [t]f'(t) dt +kf(k) - mf(m+1) \\ = -\int_{y}^x [t]f'(t) dt + \int_{k}^x [t]f'(t) dt + \int_{y}^{m+1} [t]f'(t) dt + kf(k) - mf(m+1) \\ = -\int_{y}^x [t]f'(t) dt + kf(x) - kf(k) + mf(m+1) - mf(y) + kf(k) - mf(m+1), $$

which implies

$$\tag{5}\sum_{y < n \leqslant x} f(n) = -\int_{y}^x [t]f'(t) dt + kf(x) - mf(y) $$

Now you should be able to complete the proof by subtracting (6) from (5).

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