2

$\underline{\textbf{Given:}}$

  • $m ~\in ~\mathbb{R}, \,m \not\in \{-1, 0, 1\}.$
  • $V$ is an interval in $\mathbb{R}~$ which implies that $V \subseteq \mathbb{R}.$
  • $f~$ is any generic function such that $~f ~: ~V \to \mathbb{R}.$
  • For $n \in \mathbb{Z^+}$, $~f^{(n)}(x)~$ denotes the n-th derivative of $f(x)$.
  • $f^{(1)}(x), ~f^{(2)}(x), ~f^{(3)}(x), ~$ and $~f^{(4)}(x)~$ are all well defined on $V$.
  • The explicit expression for $~f^{(2)}(x)~$ is known.

$\underline{\textbf{To calculate:}}$
$$I = \int_0^x \left[f^{(2)}(t)\right]^m \, dt \text{ or } J = \int \left[f^{(2)}(x)\right]^m \, dx.$$

For $I$, it is assumed that $[~-|x|~, ~|x|~] \subseteq V.$

For $J$, is is assumed that the function $J$ will only be applied within the interval $V$.

My intention is to find a closed form expression for either $I$ or $J$ in terms of either $f^{(2)}(x)$, or $f^{(3)}(x).~$ I do not want the closed form expression to involve the $\int$ symbol.

For example, given the alternative problem of $$\int \left[f^{(2)}(x)\right]^m ~f^{(3)}(x) ~dx,$$ the expression that I would be looking for is

$$\frac{\left[f^{(2)}(x)\right]^{(m+1)}}{m+1}.$$

Clarification:
It is to be assumed that the explicit expression for $~f^{(2)}(x)~$ is known. This means that the expression for $~f^{(3)}(x)$ can be readily calculated. This means that including an $~f^{(3)}(x)$ term in the answer is fair game.

Alternatively, although it is given (for example) that $f^{(1)}(x)~$ is well defined on $V,~$ this doesn't necessarily imply that $f^{(1)}(x)~$ can be readily calculated from the known expression $~f^{(2)}(x).~$ This means that $f^{(1)}(x)~$ can not be included in the answer.

More Clarification
In the "Helper Function" subsection of the "My Work" section of this query, I all but conclude that the query has no solution. However, I lack the mathematical knowledge or sophistication to do anything but speculate on this. Assuming that my speculation is accurate, this query is reduced to (somehow) demonstrating that there is no solution.

$\underline{\textbf{Background:}}$
This query was inspired by a very similar mathSE query that was closed. I became intrigued by the problem and unsuccessfully attacked it.

For my background, about a decade ago I survived self-studying "Calculus 2nd Ed. Volume 1" (Tom Apostol : 1966). Beyond that, my general math comprehension/knowledge is about that of an undergraduate math major.

$\underline{\textbf{Overview of My Work:}}$
The "My Work" section of the query has the following subsections:

  • Closed Form Possibilities

  • Riemann Sum

  • Differential Equations

  • Helper Function (a brute force idea of mine)

  • Reconsideration of the Helper Function

  • Substitution

  • Integration by Parts


$\underline{\textbf{My Work:}}$

$\underline{\text{Closed Form Possibilities:}}$

I give priority to this section, since the first question that should be asked is : Is this a fool's errand? Certainly, for a wide variety of explicit functions $f$, an answer can be computed. However, the question is whether there is a universal answer that will apply to any generic function $f$ that satisfies the constraints of the problem.

This question goes beyond anything discussed by Apostol. Further, I have never seen this question broached anywhere. Consequently, I have no opinion here.

$\underline{\text{Riemann Sum:}}$

$$I ~=~ \lim_{n \to \infty} \left\{~~ ~\frac{x}{n} \times \left[~~ ~\sum_{i=1}^n ~\left( ~~\left\{ f^{(2)}\left(\frac{ix}{n}\right) \right\}^m ~~\right) ~~\right] ~~\right\}.$$

I set the expression up on the off chance that it would yield insights into the problem. It hasn't triggered any ideas for me.


$\underline{\text{Differential Equations:}}$

Those experienced in DE are going to laugh at this section. I am clearly blundering in the dark. If I have my syntax right, then superficially, the problem seems to resemble a non-linear second order ordinary differential equation.

Besides the obvious problem that this is beyond the scope of Apostol's Volume 1, and seems to be beyond the scope of Apostol's Volume 2, there is a separate problem. How do I set up the functional equation?

Ignoring the fact that this is a non-linear DE, Apostol's typical second order DE looks like

$$P_0(x)y'' + P_1(x)y' + P_2(x)y = R(x).$$

It seems to me that I am searching for a function $F(x)$ that is related to the generic function $f(x)$ by the equation

$$F^{(1)}(x) = \left[f^{(2)}(x)\right]^m.$$

As near as I can figure, this means that I should set $R(x) = \left[f^{(2)}(x)\right]^m,$ set $F(x) = y$ and attack the DE

$$y' = R(x).$$ Apostol's solution to (for example) $y' + P(x)y = Q(x)$ requires that $\int Q(x) dx$ be calculated. Therefore, it seems to me that this approach accomplishes nothing.

It occurred to me to explore reversing the problem. That is, pretend that the generic $F(x)$ is known and that I am trying to solve for $f(x)$. The equation looks like

$$\left(y''\right)^m = F^{(1)}(x).$$

This (laughably) appears to have two advantages and one disadvantage.

  • Advantage - the complicated expression, $\left(y''\right)^m$ is on the LHS, which (intuitively) seems like where it belongs.
  • Advantage - if the solution requires that the right side be integrated, the integral will immediately evaluate to $F(x).$
  • Disadvantage - Apostol does not discuss non-linear second order DE's

Using Meta-Cheating to return to earth, is it just me, or is the whole idea of using DE's in this way to solve an integration problem laughable?


$\underline{\text{Helper Function:}}$

In order to understand this approach, I offer a long-winded mathSE example where I used this aproach before. In this long-winded example, you only need to examine the beginning of the Addendum through note 3 in the General Considerations subsection. The link is long-winded answer.

Transplanting this idea to the present problem, I need to find a function $~g(x)~$ so that

  • the integrand can be changed to $$ \left[ f^{(2)}(x) ~g(x) \right]^m ~\times~ \left [g(x) \right]^{(-m)} $$
  • where the following formula holds

$$ \frac{d}{dx} ~ \left[ f^{(2)}(x)~g(x) \right] ~=~ \left[g(x)\right]^{(-m)}. $$

If this is successful then $$ J ~=~ \frac{1}{m+1}~~ \left[ f^{(2)}(x) ~g(x) \right]^{(m+1)}. $$

In searching for $~g(x),~$ I note that $$ \frac{d}{dx} ~ \left[ f^{(2)}(x)~g(x) \right] ~=~ \left[f^{(2)}(x) ~g^{(1)}(x)\right] ~+~ \left[f^{(3)}(x) ~g(x)\right]. $$

Therefore $$ \left[g(x)\right]^{(-m)}~ \text{must equal} ~\left[f^{(2)}(x) ~g^{(1)}(x)\right] ~+~ \left[f^{(3)}(x) ~g(x)\right]. \tag{1} $$

Based on equation (1), the natural first try for specifying $g(x)$ seems to be

$$g(x) ~=~ a \times \left[f^{(2)}(x)\right]^b ~\text{where} ~~a,b ~\in ~\mathbb{R}. $$

This gives $$f^{(2)}(x) \times g(x) ~= a \times \left[f^{(2)}(x)\right]^{(b+1)}. $$

This implies that $$\frac{d}{dx} ~ \left[ f^{(2)}(x)~g(x) \right] ~=~ (b+1) (a) \times \left[f^{(2)}(x)\right]^b \times ~f^{(3)}(x). \tag{2} $$

Equation 1 now requires that

$$ \frac{1}{a^m \times \left[f^{(2)}(x)\right]^{(bm)}} ~= ~\text{RHS of (2)}. \tag{3} $$

This implies that

$$ (b+1) (a)^{(m+1)} \times \left[f^{(2)}(x)\right]^{(bm + b)} \times ~f^{(3)}(x) ~=~ 1. \tag{4} $$

In effect, the requirement is that scalars $~a~$ and $~b~$ be found to force a relationship like $$ C_1 \times \left[f^{(2)}(x)\right]^{-C_2(m + 1)} ~=~ f^{(3)}(x). $$ where $~C_1~$ and $~C_2~$ are scalars.

Since the underlying function, $~f(x),~$ is a generic function, no such possible relationship can be presumed to be possible for all $~x~$ throughout the interval $~V$.

For the same reason, formatting $~g(x)~$ as

$$\left\{a \times \left[f^{(2)}(x)\right]^b\right\} ~+~ \left\{c \times \left[f^{(2)}(x)\right]^d\right\} \tag{5}$$

must also fail. The $~f^{(3)}(x)~$ factor will appear in $$\frac{d}{dx} ~ \left[ f^{(2)}(x)~g(x) \right]$$

but will not appear in $~g(x)^{(-m)}.$

An identical consideration with respect to $~f^{(4)}(x)~$ prevents $~g(x)~$ from being formatted as

$$a \times \left[f^{(2)}(x)\right]^b \times \left[f^{(3)}(x)\right]^c. \tag{6}$$

Consequently, unless I am overlooking some offbeat attempt, this approach will not work. Since this approach is based on brute force, this strongly suggests (to me) that there is no solution to the query.


$\underline{\text{Reconsideration of the Helper Function:}}$

It could be argued that although $~f(x)~$ is deemed to be a generic function, both $~f^{(2)}(x)~$ and $~f^{(3)}(x)~$ are deemed to be known functions.

Therefore, it can be argued that the query is actually asking for an algorithm to be used when computing $~\int ~[f^{(2)}(x)]^m~dx.~$ The brute force idea from the previous section does represent such an algorithm.

That is, a first step in the algorithm could be to check whether scalars $~a~$ and $~b~$ can be found so that equation (4) from the previous section is satisfied, for all $~x \in V.~$ Assuming that such a relationship can't be satisfied for the specific scalar $~m~$ and the specific functions $f^{(2)}(x)~$ and $~f^{(3)}(x)~$ being attacked, then the algorithm might then move on to the attempts represented by equations 5 and 6 in the previous section.

I am specifically excluding this interpretation of my query since it would render my query all but meaningless. Therefore, the requirement is that $I$ or $J$ be blindly computed in the same way that one could blindly compute

$$\int \left\{\left[f^{(2)}(x)\right]^m ~f^{(3)}(x) \right\}~dx ~~~~\text{as}~~~~ \frac{\left[f^{(2)}(x)\right]^{(m+1)}}{m+1}.$$


$\underline{\text{Substitution:}}$

Although my instinct is that there is an underlying association between this strategy and the strategy employed in the "Helper Function" section, I will go ahead and work this section independently.

The first try is $$t = f^{(2)}(x).$$

Under this approach you have the informal equation

$$d \left[f^{(2)}(x)\right] = f^{(3)}(x) dx = dt \implies dx = \frac{dt}{f^{(3)}(x)}.$$

This means that the integral $J$ is partially converted to

$$\int t^m \frac{dt}{f^{(3)}(x)}.$$

This approach has the same problem as the approach taken in the "Helper Function" subsection. That is, since $f(x)$ is a generic function, $f^{(3)}(x)$ can't be expressed as a function of $t$.

Further, I think that the same general discussion in the "Helper Function" subsection will pertain to miscellaneous other approaches taken here. Therefore, it seems to me that if the methods employed in the "Helper Function" subsection must fail, then this approach must also fail.


$\underline{\text{Integration by Parts:}}$

$$ \begin{array}{| r r l r | r r l |} \hline r & = & ~~~~\frac{1}{f^{(3)}(x)} & ~~~ & s & = & \frac{\left[f^{(2)}(x)\right]^{(m+1)}}{m+1} \\[8pt] \hline \\[1pt] dr & = & -\frac{f^{(4)}(x)}{\left[f^{(3)}(x)\right]^2} ~dx& ~~~ & ds & = & \left[f^{(2)}(x)\right]^{m} ~f^{(3)}(x) ~dx \\[8pt] \hline \end{array} $$ As in previous subsections, this is just "chasing my tail".

K.defaoite
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