Where $f$ is a function defined in $\mathbb{R}$ with countinuos derivative in all $\mathbb{R}$, for each $n\in \mathbb{N}$ and the function $\lfloor x \rfloor$ is the floor function.
I tried using integration by parts:
I have
$$\int_1^n f'(x)\lfloor x \rfloor dx + \int_1^n \lfloor x \rfloor ' f(x) dx = \lfloor n \rfloor f(n)-\lfloor 1 \rfloor f(1)$$
Then $$\int_1^nf(x)\lfloor x \rfloor ' dx + \lfloor 1 \rfloor f(1) = \lfloor x \rfloor f(n)- \int_1^n \lfloor x \rfloor f'(x) dx$$
But I have(Is this correct?):
$$\int_1^nf'(x) \lfloor x \rfloor dx= \sum_{i=1}^n \lfloor x \rfloor \int_i^{i+1}f'(x) dx= \sum_{{i = 1}}^{n} f(i)$$
So now I end with:
$$\sum_{{i = 1}}^{n} f(i)= \lfloor x \rfloor f(n) -\int_1^nf(x)\lfloor x \rfloor ' dx - \lfloor 1 \rfloor f(1) $$ But I don't know how to proceed from here.
I see this problem in the book Real Analysis from Carothers 14.37.a
HINT: note that $\lfloor x\rfloor\le x$, and that $\lfloor x\rfloor$ is a constant function in $[n,n+1)$ for any chosen $n\in\Bbb N$, so
$$\int_1^n f'(x)\lfloor x\rfloor\, dx\\=\int_1^2 f'(x)\cdot 1\, dx+\int_2^3 f'(x)\cdot 2\, dx+\ldots+\int_{n-1}^n f'(x)(n-1)\, dx\\ =\sum_{k=1}^{n-1}k\int_k^{k+1}f'(x)\, dx\tag1$$
Now applying the fundamental theorem of calculus we find that
$$\sum_{k=1}^{n-1}k\int_k^{k+1}f'(x)\, dx=\sum_{k=1}^{n-1}k(f(k+1)-f(k))=\sum_{k=1}^{n-1} kf(k+1)-\sum_{k=1}^{n-1}kf(k)\\ =\sum_{k=2}^n (k-1)f(k)-\sum_{k=1}^{n-1}kf(k)=(n-1)f(n)-f(1)-\sum_{k=2}^{n-1}f(k)$$
Can you conclude from here, right?