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Euler's summation by parts formula states that: $$ \sum_{y < n \leq x} f(n) = \int_y^x{f(t)dt} + \int_y^x(t - \lfloor t \rfloor)f'(t)dt +f(x)(\lfloor x \rfloor - x) -f(y)(\lfloor y \rfloor -y)$$ (source: https://asgarli.files.wordpress.com/2013/02/euler-summation-formula.pdf)

Lets only look at the first of those two terms (for the time being):

$$\int_y^x{f(t)dt} + \int_y^x(t - \lfloor t \rfloor)f'(t)dt$$

$$\int_y^x(t - \lfloor t \rfloor)f'(t)dt = \int_y^xtf'(t)dt - \int_y^x\lfloor t \rfloor f'(t)dt$$

Taking $\int_2^xtf'(t)dt$ we have through integration by parts, where $u=t$ and $f'(t)=dv$

$$\int_y^xtf'(t)dt = tf(t)\Big|_y^x - \int_y^xf(t)dt = xf(x) - yf(y) - \int_y^xf(t)$$ Substituting into our original equation we have: $$ \sum_{y < n \leq x} f(n) = \\\int_y^x{f(t)dt} + \int_y^x(t - \lfloor t \rfloor)f'(t)dt +f(x)(\lfloor x \rfloor - x) -f(y)(\lfloor y \rfloor -y) = \\\int_y^x{f(t)dt} + xf(x) - yf(y) - \int_y^xf(t)dt - \int_y^x\lfloor t \rfloor f'(t)dt+f(x)(\lfloor x \rfloor - x) -f(y)(\lfloor y \rfloor -y) =\\xf(x)-yf(y)+(\lfloor x \rfloor - x)f(x) -f(y)(\lfloor y \rfloor -y) -\int_y^x\lfloor t \rfloor f'(t)dt= \\f(x)(x+\lfloor x \rfloor -x) - f(y)(y+\lfloor y \rfloor -y) - \int_y^x\lfloor t \rfloor f'(t)dt= \lfloor x \rfloor f(x) - \lfloor y \rfloor f(y) - \int_y^x\lfloor t \rfloor f'(t)dt$$

This result is obviously incorrect (it is significantly shorter and simplet than the original formula for one), also, take, for instance $f(n) = 1$, $y=1$, $x=2$

I particularly find it suspicious that the first term is cancelled out in its entirety by the second. But I can not seem to understand where my flaw is. Any help would be much appreciated.

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    Which summation formula? – 2017-01-21
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    https://asgarli.files.wordpress.com/2013/02/euler-summation-formula.pdf – 2017-01-21
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    Thanks, I just ask because there is more than one summation formula named after Euler. :-) – 2017-01-21
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    all good :) Thanks for clarifying... something interesting to add to my google backlog :) – 2017-01-21
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    The sum is $$\sum_{y \color{red}{<} n \leq x}f(n).$$ – 2017-01-21
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    edited. Good spot – 2017-01-21

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It appears the function is completely correct. In fact, in the link you gave, they derive the exact same expression, a few lines from the bottom. And if you try your example, there is only one term in the sum, and it equals $1$, whereas on the other side of the equation, we get $2f(2)-f(1)=1$, which is exactly as we wanted. What's the problem?

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    Interesting I didn't actually read that source very closely I've been following Apostol. The difference is that in that source the result is equal to the integral, while in my calculations the result is equal to the whole sum... – 2017-01-21
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    @AbrahamP I'm looking at the block of equations right after "Therefore". It certainly looks like they equate the sum to your expression. – 2017-01-21
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    Oh.. yeah I'm blind... That's a...surprising result somehow... but yeah I suppose there's nothing off :/ Thanks for you help! – 2017-01-21