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Is there an integral form equivalant to $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$

e.g. do m,n exists such that for any f following will hold true : $\int\limits_{t=0}^x {\big(f(t)\big)^m} \operatorname dt = \bigg(\int\limits_{t=0}^x f(t) \operatorname dt\bigg)^n$

jimjim
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    Doesn't $f(t)=t$ work with $m=3$, $n=2$? both yielding $x^4/4$. – Sungjin Kim Aug 18 '16 at 00:10
  • If you want to find $(m,n)$ s.t. the equation holds for any $f$, then it is impossible. By considering $f(x)$ and $2f(x)$, it gives $m=n$ which doesn't holds for $f(t)=t$. – Seewoo Lee Aug 18 '16 at 00:13

2 Answers2

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The analogue is exactly what you'd expect: $$ \int_0^xt^3\;dt=\frac{x^4}{4}=\Big(\int_0^xt\;dt\Big)^2$$

This is not a surprise because $\frac{1}{n^4}\sum_{k=1}^{n}k^3$ and $\frac{1}{n^2}\sum_{k=1}^nk$ are Riemann sums for $\int_0^1t^3\;dt$ and $\int_0^1t\;dt$ respectively.

carmichael561
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The number-theoretic function $f: \mathbb{N} \to \mathbb{R}$, $f(k) = k$, is just a discrete analogue of the identity $f(x) = x$. So try it out: if we have such an $n,m$, $$\frac{1}{m+1}x^{m+1} = \frac{1}{2^n}x^{n}$$ Thus $m+1=2^n$, $n = m+1$, and looking for integer solutions we get $m=3$, $n=2$. Thus, $$\int_0^x t^3 dt = \Big ( \int_0^x t dt \Big )^2$$

GiantTortoise1729
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