Let $f(u)$ be a continuous function and, for any real number $u$, let $[u]$ denote the greatest integer less than or equal to $u$. Show that for any $x \gt 1$, $$\int \limits_1 ^x [u]([u]+1)f(u)du=2\sum \limits_{i=1}^{[x]}i \int \limits_i ^x f(u)du $$
1 Answer
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$\int_{1}^{x} [u]([u]+1) f(u) du = 1 \cdot 2 \cdot \int_{1}^{2} f(u) du + 2 \cdot 3 \int_{2}^{3} f(u) du + \dots + [x]\cdot ([x]+1) \int_{[x]}^x f(u) du = 2 \cdot (\int_{1}^{2} f(u) du + \int_{2}^{3} f(u) du + \int_{3}^{4} f(u) du + \dots + \int_{[x]}^{x} f(u) du) + \\ 4 \cdot (\int_{2}^{3} f(u) du + \int_{3}^{4} f(u) du + \int_{4}^{5} f(u) du + \dots + \int_{[x]}^{x} f(u) du)+ \\ 6 \cdot (\int_{3}^{4} f(u) du + \int_{4}^{5} f(u) du + \int_{5}^{6} f(u) du + \dots + \int_{[x]}^{x} f(u) du)+\\ \vdots \\ +2 \cdot [x] \cdot \int_{[x]}^x f(u) du = RHS $
