8

For a function $f$ I know that: $$\int{f'(r)dr}=f(r)$$ where $f(r)$ is known. knowing the result of this integral how can i calculate $$\int{(f'(r))^2dr}$$ Is there any relation between these integrals?

Fermat
  • 5,230
Pouya
  • 145
  • 2
    If you know the second derivative you can solve it by integrating by parts. – David H Jun 01 '14 at 20:32
  • 2
    No, there is no general formula involving only $f$, $f'$ and $r$ for this integral. It might be a nice exercise to try to prove this. – Winther Jun 01 '14 at 20:35
  • I don't think there's any meaningful relation. Just think of $;f(x)=\log x;,;;f'(x)=\frac1x;$ ... – DonAntonio Jun 01 '14 at 20:37

2 Answers2

6

While there is no explicit formula of the exact kind you desire, if one is willing to reparametrize, there is an integral-free parametrization of the curves of the form $\bigl(x,f(x),g(x)\bigr)$ for which $g'(x) = (f'(x))^2$: As long as $f''(x)$ is non-vanishing, one can reparametrize such a curve in the form $$ \bigl(x,f(x),g(x)\bigr) = \bigl(u''(t),\ t\,u''(t){-}u'(t),\ t^2u''(t){-}2tu'(t){+}2u(t)\,\bigr) $$ where $u$ is a function of $t = f'(x)$. Conversely, if $u$ is an arbitrary function of $t$ for which $u'''(t)$ is nonvanishing, the curve on the right hand side of the equation is always of the form $\bigl(x,f(x),g(x)\bigr)$ for some functions $f$ and $g$ satisfying $g'(x) = (f'(x))^2$.

Perhaps this is the sort of thing that one can use when one is trying to get an integral-free description of the relationship between the two functions $f$ and $g$ (and $x$, of course).

Robert Bryant
  • 1,156
  • Hi, I don't have the math background to understand what to make of the curve you've described. Do you have any suggestions as to good background material? This kind of seems like differential geometry, is that right? – John Madden May 16 '21 at 17:56
  • @JohnMadden: Yes, it's from differential geometry, but you don't need any differential geometry to see that with $\bigl(x,f(x),g(x)\bigr)$ as given, we have $$\frac{\mathrm{d}g}{\mathrm{d}x} = t^2 = \left(\frac{\mathrm{d}f}{\mathrm{d}x} \right)^2,$$ just the Chain Rule. – Robert Bryant May 17 '21 at 19:39
  • Thanks for this helpful formula! Where did this come from, and do you know of a systematic way of generating these sorts of formulas? It's not completely clear to me how this was found... – ccbreen May 19 '21 at 05:52
  • 2
    @ccbreen: Yes, there is a systematic way to search for such 'integral-free' formulae for curves. This is known in the control theory literature as 'differential flatness'. The test for such in general is too involved to write out here, but the idea is that you want to find coordinates in which the differential system $$ df - p,dx = dg - p^2,dx = 0$$ can be written in "Engel normal form", i.e. $$ du - v,dt = dv - w,dt = 0,$$ whose integral curves can be written in the form $(t,u,v,w) = \bigl(t,h(t),h'(t),h''(t)\bigr)$. Engel's theorem ensures that you can do this. – Robert Bryant May 19 '21 at 07:32
  • @RobertBryant, could you possibly comment on this: https://math.stackexchange.com/questions/4771881/weak-version-of-maximum-principle-for-not-quite-subharmonic-functions ? – Mikhail Katz Sep 19 '23 at 16:27
2

Let $f(x)=e^{x^2}$, so $(f'(x))^2=4x^2e^{2x^2}$. But $\int4x^2e^{2x^2}\,dx$ can't be evaluated in terms of elementary functions (exponentials, trig functions, polynomials, $n$th roots, etc.).

Gerry Myerson
  • 179,216