How do I prove for a continuous function $f$ that
$$ \int_1^x\lfloor{u}\rfloor\lfloor{u+1}\rfloor f(u)du=2\sum_{i=1}^{\lfloor{x}\rfloor}i\int_i^xf(u)du\ ? $$ This is what I've come up with so far: $$ \int_1^x\lfloor{u}\rfloor\lfloor{u+1}\rfloor f(u)du=\sum_{i=1}^{\lfloor{x}\rfloor-1}i(i+1)\int_i^{i+1} f(u) du+\lfloor x\rfloor \lfloor x+1\rfloor \int_{\lfloor x\rfloor}^{x}f(u)du. $$ It doesn't seem to be of any help though.
A strange-looking identity
3
$\begingroup$
calculus
integration
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0Have you tried sketching what the function $\lfloor u \rfloor\lfloor u+1 \rfloor$ looks like, say for $1 \leq u \leq x = 2\pi$ just to pick a number? It will give you a hint about the value of the integral more easily than mathematical manipulations that are not getting you anywhere – 2012-05-12
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0I did. I still can't figure out anything. http://tinyurl.com/floor-plot – 2012-05-12
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0Got something from an answer below? – 2012-07-25