Any way to simplify $int sqrt{f'(x)} mathrm{d} x$?
$begingroup$
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.
calculus integration
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
$endgroup$
|
show 2 more comments
$begingroup$
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.
calculus integration
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
$endgroup$
$begingroup$
Please make the question more specific. Otherwise I can just say $intsqrt{f'(x)}dx$ 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?
$endgroup$
– YiFan
Dec 14 '18 at 0:00
$begingroup$
@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.
$endgroup$
– 0x539
Dec 14 '18 at 0:02
$begingroup$
So the simplification is $intsqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify.
$endgroup$
– YiFan
Dec 14 '18 at 0:04
$begingroup$
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.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 0:05
1
$begingroup$
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.
$endgroup$
– GEdgar
Dec 14 '18 at 1:14
|
show 2 more comments
$begingroup$
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.
calculus integration
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
$endgroup$
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.
calculus integration
calculus integration
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
edited Dec 13 '18 at 23:47
0x539
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
asked Dec 13 '18 at 23:33
0x5390x539
1,067317
1,067317
$begingroup$
Please make the question more specific. Otherwise I can just say $intsqrt{f'(x)}dx$ 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?
$endgroup$
– YiFan
Dec 14 '18 at 0:00
$begingroup$
@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.
$endgroup$
– 0x539
Dec 14 '18 at 0:02
$begingroup$
So the simplification is $intsqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify.
$endgroup$
– YiFan
Dec 14 '18 at 0:04
$begingroup$
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.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 0:05
1
$begingroup$
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.
$endgroup$
– GEdgar
Dec 14 '18 at 1:14
|
show 2 more comments
$begingroup$
Please make the question more specific. Otherwise I can just say $intsqrt{f'(x)}dx$ 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?
$endgroup$
– YiFan
Dec 14 '18 at 0:00
$begingroup$
@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.
$endgroup$
– 0x539
Dec 14 '18 at 0:02
$begingroup$
So the simplification is $intsqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify.
$endgroup$
– YiFan
Dec 14 '18 at 0:04
$begingroup$
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.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 0:05
1
$begingroup$
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.
$endgroup$
– GEdgar
Dec 14 '18 at 1:14
$begingroup$
Please make the question more specific. Otherwise I can just say $intsqrt{f'(x)}dx$ 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?
$endgroup$
– YiFan
Dec 14 '18 at 0:00
$begingroup$
Please make the question more specific. Otherwise I can just say $intsqrt{f'(x)}dx$ 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?
$endgroup$
– YiFan
Dec 14 '18 at 0:00
$begingroup$
@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.
$endgroup$
– 0x539
Dec 14 '18 at 0:02
$begingroup$
@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.
$endgroup$
– 0x539
Dec 14 '18 at 0:02
$begingroup$
So the simplification is $intsqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify.
$endgroup$
– YiFan
Dec 14 '18 at 0:04
$begingroup$
So the simplification is $intsqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify.
$endgroup$
– YiFan
Dec 14 '18 at 0:04
$begingroup$
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.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 0:05
$begingroup$
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.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 0:05
1
1
$begingroup$
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.
$endgroup$
– GEdgar
Dec 14 '18 at 1:14
$begingroup$
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.
$endgroup$
– GEdgar
Dec 14 '18 at 1:14
|
show 2 more comments
1 Answer
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$begingroup$
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.
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
$endgroup$
add a comment |
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$begingroup$
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.
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
$endgroup$
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.
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
answered Dec 14 '18 at 12:59
George CooteGeorge Coote
855311
855311
add a comment |
add a comment |
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$begingroup$
Please make the question more specific. Otherwise I can just say $intsqrt{f'(x)}dx$ 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?
$endgroup$
– YiFan
Dec 14 '18 at 0:00
$begingroup$
@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.
$endgroup$
– 0x539
Dec 14 '18 at 0:02
$begingroup$
So the simplification is $intsqrt{f'(x)}dx$. Done! Unless this expression is not actually allowed, in which case you must specify.
$endgroup$
– YiFan
Dec 14 '18 at 0:04
$begingroup$
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.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 0:05
1
$begingroup$
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.
$endgroup$
– GEdgar
Dec 14 '18 at 1:14