I'd like to know if

$X_3 = \dfrac{1}{\tau} \int_0^\tau f^3(t) dt$

can be expressed in terms of

$X_1 = \dfrac{1}{\tau} \int_0^\tau f(t) dt$


$\dfrac{1}{\tau} \int_0^\tau f^2(t) dt$.

If not possible, is there any good approximation to $X_3$ assuming that both $\lvert f(t) \lvert < c_1$ and $\lvert \int_0^\tau f(t) dt \lvert < c_2$ for some finite positive constants which means that $X_1 \approx 0$ and $\lim_{\tau \to \inf} X_1(\tau) = 0$


1 Answer 1


$$ X_3 = \frac{1}{\tau} \int_0^\tau \big( \frac{d}{dt}t \, X_1(t) \big)^3 dt $$ here by FTC $\frac{d}{d\tau}\tau X_1(\tau) = f(\tau)$.

  • $\begingroup$ Are you sure about that? Because it seems to me that, $\dfrac{ }{d\tau] (\tau X_1(\tau)) = f(\tau) + X_1(tau)$. $\endgroup$ Sep 1, 2016 at 6:18
  • $\begingroup$ I think I am sure. $\tau X_1(\tau) = \int_0^\tau f(t)dt$ therefore $\frac{d}{d\tau}\tau X_1(\tau) = \frac{d}{d\tau}\int_0^\tau f(t)dt = f(\tau)$ $\endgroup$
    – iamvegan
    Sep 1, 2016 at 9:16

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