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Trying to calculate the length of some geodesic I came across an expression of the form $$ \int \sqrt{f'(x)} \,\mathrm{d}x $$ which got me wondering if there is any way to simplify or calculate such integrals in general (where $f: \mathbb{R} \to \mathbb{R}$ is sufficiently regular).

EDIT: I'm looking for a simplification in terms on $f$. (Anti-)derivatives of $f$ and integral transforms of $f$ would be interesting too.

EDIT 2: The asymptotic expansion for the integral of a product $$ \int f g = \sum_{k=0}^{\infty} (-1)^k f^{(k)} g^{(-1-k)} $$

@AmbretteOrrisey has given in the comments actually looks quite interesting (even if it is not very useful). I guess something similar for my expression would be nice, too.

Carlos Esparza
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  • Please make the question more specific. Otherwise I can just say $\int\sqrt{f'(x)}dx$ and be done with it. Are you specifying that the answer can only be in terms of $f(x)$, $f'(x)$ and $\int f(x)dx$? What about $f(x^2)$ or $f''(x)$? Do you allow those? – YiFan Tey Dec 14 '18 at 00:00
  • @YiFan As I wrote in my edit, those are all fine. Basically you know all values of $f$ as well as of its derivatives and integrals. – Carlos Esparza Dec 14 '18 at 00:02
  • So the simplification is $\int\sqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify. – YiFan Tey Dec 14 '18 at 00:04
  • I feel like this would be a hell of a lot easier if we knew $f$ or $f'$. Like I get you want a general statement but I don't know of anything that would apply. – PrincessEev Dec 14 '18 at 00:05
  • @EeveeTrainer I'm no longer interested in the specific integral that gave me this idea, so there only is the general case. – Carlos Esparza Dec 14 '18 at 00:24
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    No, there is in general no simple form for this. It is a frustration of calculus teachers that there is only a short list of examples where the arc length of $y=f(x)$ can be computed explicitly. Which is why all the calculus books have the same exercises for this. – GEdgar Dec 14 '18 at 01:14
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    Not likely - you generally don't get 'pleasant' theorems for integration of functions of general functions in the way you do for differentiation. There might conceivably be some kind of infinite series that produces a tractable asymptotic expansion in certain cases - as there is for the product of two functions - $\sum_{k=0}^\infty(-1)^k\operatorname{f^{(k)}}\operatorname{g^{(-1-k)}}$ ... but the way these integrals are generally done is on an ad hoc basis - broaching the properties of whatever particular function is the argument of the square-root function. – AmbretteOrrisey Dec 14 '18 at 14:17

1 Answers1

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I think you're looking for some sort of analog to $$\frac {\mathrm d(\sqrt {f'(x)})} {\mathrm dx} = \frac {f''(x)} {2 \sqrt{f'(x)}} $$

But for $\int \sqrt{f'(x)} \mathrm dx$. As alluded to in the comment, you will probably be disappointed to know that there is no such connection. Unfortunately integrals rarely work out this nicely.

Indeed, $\int \sqrt{f(x)} \mathrm dx$ may be completely different from $\int f(x) \mathrm dx$. For example:

$$\int (1 - x^3) \mathrm dx$$

is elementary and a polynomial itself, but:

$$\int \sqrt{1 - x^3} \mathrm dx$$

can only be resolved in terms of the elliptic integrals. (but has nice definite integrals over eg. $[0,1]$ with the aid of the $\Gamma$ function)

It ultimately boils down to the properties of $f'$. Certain $f'$s will resolve nicely, for example squares of functions with known antiderivatives, or in general even powers of nice functions, (eg. $\sqrt{f'(x)} = \text{polynomial} \cdot \sin x$ which will fall out, eventually, by parts) others will have unwieldly elementary antiderivatives, (eg. reciprocals of very high-order polynomials, which will eventually decompose into partial fractions and admit elementary anti-derivatives) or none at all (eg. $\frac {\sin x} x$).

Predicting whether a function has an elementary antiderivative is a question in differential algebra and is not an easy question at all. I don't have the background to understand it, but there is an answer here that addresses the topic.

George C
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