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How would I compute the integral $\int{\sqrt{dx}}$?

I would suppose it is recursive like this:

$\int{(dx)^2} = (x + C)dx$

Is this a well formed mathematical term? I can imagine it is because

$\iint{(dx)^2} = \frac{x^2}{2} + Cx + D$ and thus $\frac{d\iint{dx^2}}{dx} = x + C$ and then $\int{dx^2} = (x + C)dx$.

WorldSEnder
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    It is just $\sqrt x + C$ – Ali Caglayan Dec 11 '14 at 08:55
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    You can interpret '$d$' and '$\int$' as operators on functions. In that context $\int d(f(x))=f(x)$ – drhab Dec 11 '14 at 08:56
  • $df(x) = f^\prime(x) dx$. So $\int df = \int f^\prime(x) dx = f+C$ and the result from the two earlier comments is correct. – Thomas Dec 11 '14 at 09:11
  • I corrected the question. It was not really what I meant, I forgot some brackets – WorldSEnder Dec 11 '14 at 09:11
  • What is $C$ here? It can stand for any constant. This way $\int$ is sending functions to a set of functions. The function $x^2$ is one of them. Then $\int$ is not an operator on functions anymore and you have stepped out of context I mentioned. – drhab Dec 11 '14 at 09:14
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    In what context have you seen this notation ($\sqrt{dx}$)? I have never seen it before. – mrf Dec 11 '14 at 09:14
  • It was something that appeared to me because (dx)^2 is well formed also. Why should I not apply the sqrt-function to dx? – WorldSEnder Dec 11 '14 at 09:15
  • Your edit makes my comments irrelevant. Btw, just like @mrf I am not familiar with the updated notation. – drhab Dec 11 '14 at 09:16
  • If you just made it up yourself, it's up to you to tell others how it is supposed to be interpreted. Preferably with a useful interpretation. – mrf Dec 11 '14 at 09:17
  • It means what it states. You just substitute the definition of dx into $\sqrt(dx)$. I don't think that needs a definition. – WorldSEnder Dec 11 '14 at 09:20
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    What do you mean by "$dx$" then? Is it a differential 1-form? Something else? Do you intend the symbol $\sqrt{dx}$ to have meaning on its own? Or just coupled with an integral sign? [In other words: do you see where I'm heading? Of course it needs a definition if anyone is supposed to make sense out of it.] – mrf Dec 11 '14 at 09:30
  • http://math.stackexchange.com/questions/143222/what-does-dx-mean – Bumblebee Dec 11 '14 at 10:59
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    @WorldSEnder, how about if I substitute the definition (from logic) of $\implies$ into $\sqrt\implies$. Does that not need a definition? – Barry Cipra Dec 11 '14 at 15:25

1 Answers1

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$\int{\sqrt{dx}}$ is an ambiguous symbolism, as already pointed out by several posters.

The tone of the question draw me to think that you are looking for something like "fractionnal calculus", i.e. antiderivative and derivative of non-integer degree. In this sense, it should be better to raise the question on this form :

How would I compute $\frac{d^\nu}{dx^\nu}f(x)$ ? where $\nu=-\frac{1}{2}$ and $f(x)=1$ in your example.

For information, see:

http://mathworld.wolfram.com/FractionalDerivative.html http://mathworld.wolfram.com/FractionalIntegral.html

Your particular case corresponds to the definition of "semi-integral", i.e. fractionnal integral of order 1/2 , as shown here :

http://mathworld.wolfram.com/Semi-Integral.html

The result is : $$\frac{d^{-1/2}}{dx^{-1/2}}(1)=2\sqrt{\frac{x}{\pi}}$$

A paper for general public is available here :

https://fr.scribd.com/doc/14686539/The-Fractional-Derivation-La-derivation-fractionnaire

Note that your writting :

$\iint{(dx)^2} = \frac{x^2}{2} + Cx + D$ and thus $\frac{d\iint{dx^2}}{dx} = x + C$ and then $\int{dx^2} = (x + C)dx$

is not correct for two reasons : First, $\frac{d\iint{dx^2}}{dx}$ is not equal to $\int{dx^2}$ but is equal to $\int{dx}$. Second : The right term cannot be an infinitesimal such as $(x + C)dx$. The $dx$ is too much.

JJacquelin
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