I Discovered something crazy

1. May 25, 2014

guicortei

Guys... I'm not a mathematician, so, sorry about my informality in math...

here is something crazy that I discovered about 2 years ago.

It is a way to discover some integrals without integrate the f(x)...
It is a way knowing the f-1(x)...

there is a crazy K (constant i guess)... I really would apreciate your comments!

I attached it here

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2. May 25, 2014

micromass

Staff Emeritus
You will likely by interested in the following: http://en.wikipedia.org/wiki/Young's_inequality#Standard_version_for_increasing_functions

This is your starting equality with $a = 0$, $f(a) = 0$. The first thing to note is that the wiki page says this is true only for increasing functions. Your visual "proof" also assumes an increasing function. I don't think it holds in general, no?

Then, you should really change your notations. Something like

$$\left[\int f(x)dx\right]_{x=x} = \int f(x)dx$$

is really meaningless to me. I don't even know what you mean with the left-hand side. I think I know what you mean, it's just calculating the integral and then substituting in $x$. But this is meaningless. The reason is that the indefinite integral is not uniquely determined, but is only determined up to a constant. So something like

$$\left[\int xdx\right]_{x=x}$$

could be equal to both $x^2/2$ as $x^2/2 + 1$. So the result is ambiguous. You should really clarify what you mean.

Now, if you take $f(0)=0$, then you can always use

$$\int_0^x f(t)dt$$

This is not ambiguous. But, you really can't say that this is equal to $\int f(x)dx$. An indefinite integral and a definite integral are two very different animals. An indefinite integral is only determined up to a constant, a definite integral isn't.

So you should start by rewriting everything using a better notation. That is, a notation like

$$\left[\int f(x)dx\right]_{x=a}$$

should not be used. And you should also not use indefinite integrals $\int f(x)dx$ in this context.

3. May 25, 2014

maajdl

Your explanation is not clear, and the derivation is extremely long.
However, I think I can guess what you are asking.
I guess that your "end result" is right and extremely simple to derive.
It is just a consequence if integration by part:

f dx = d(f x) -x df

and considering f a function of x or x a function of f.

f(x) dx(f) = d( f(x) x(f) ) -x(f) df(x)

The rest is just about trying to use clear notations.
The result is not more useful than the integration by part itself, since it just re-writing it in another way.

Last edited: May 25, 2014
4. May 25, 2014

guicortei

micromass...ok, how should I write "Integral of f(x)dx calculated on x=a"?

micromass... I tried to find the K to the function f(x)=a Ln(b x + c) + d, and I found a K=c/b(d-a), and it is impossible to assume x=0 like you said above..

I'm thinking that the K can show the form how you could write the integral of f(x) in terms of f(x), like torricelli

5. May 25, 2014

micromass

Staff Emeritus
Well, what do you mean by this?

6. May 25, 2014

guicortei

when we integrate f(x) from x=0 to x=1, the result is [the integral of f(x)] calculated with x=1 minus [the integral of f(x)] calculated with x = 0...

then... I tried to mean... integral f(x) dx from "a" to "x" is equal to [integral of f(x)](undefined) minus [integral of f(x)](undefined) with x equal "a"

7. May 25, 2014

micromass

Staff Emeritus
OK, but do you understand why "[the integral of x] calculated with x=1" is meaningless on its own?

For example, what would

$$\left[\int xdx\right]_{x=1}$$

be?

Well, we can say that $x^2/2$ is a primitive, so we can substitute $x=1$ to get $1/2$.
But we might as well take the primitie $x^2 /2 + 1$. This is another perfectly allowed primitive. We substitute in $x=1$ and we get $3/2$.

The problem is that there is not a unique primitive function of $f(x)=x$ and of any other function. So the thing you want to define isn't uniquely defined.

8. May 26, 2014

guicortei

it is not me meaningless...

integral of x dx = x^2/2 + Constant

[integral of x dx](x=1) = 1/2 + Constant
[integral of x dx](x=0) = 0 + Constant (same constant)

Integral of x dx, x=0 to x=1.... is equal... [integral of x dx](x=1) minus [integral of x dx](x=0)

is equal
(1/2 + Constant) minus (0 + Constant) = 1/2

9. May 26, 2014

abitslow

micromass is very patient. why do you assume that f⁻¹ exists? For most functions, it does not.
Do NOT use x when you mean a specific point on the x-axis. Restate your entire argument, but pick a letter, preferably from early in the alphabet, say c or k.
micromass has clearly stated, but you don't seem to be listening, that a definite integral is logically very different from an indefinite integral. Do NOT conflate them.
You also seem to believe that for any function, that there is always a point c, where c = f(c). y= x+k where k≠0
comes to mind...where does y=x? { based on your terrible notation x=f(x) }
My comments are much the same as his: your notation needs to be clearly defined, which it is not AND needs to be consistent with what the various types of integrals are, which it is not. The fact that you use ambiguous notation and think it leads you anywhere is basically telling us you hear voices in your head. Crazy? It is wishful thinking.
A stopped watch is right twice a day. (Meaning an argument can occassionally work even if it is generally wrong.)

10. May 26, 2014

micromass

Staff Emeritus
I agree that $\int_0^1 xdx$ is perfectly fine since the constants drop. That's not the problem. But something like

$$\left[\int xdx\right]_{x=0}$$

is ambiguous and undefined. It depends on an unspecified constant. So it is not clear which real number you mean.

Sure, something like

$$\left[\int xdx\right]_{x=1}-\left[\int xdx\right]_{x=0}$$

is fine since the constants cancel eachother out. But any other use of the notation

$$\left[\int xdx\right]_{x=0}$$

is wrong. So you can only use the above notation when you substract the same thing (but in another constant) later on. But then your notation is nothing different than $\int_0^1 xdx$.

11. May 26, 2014

micromass

Staff Emeritus

12. May 26, 2014

maajdl

After all, you have simply rediscovered the integration by part, that can be illustrated graphically as shown on the picture below.
This picture also illustrates what you need to take care of: the picture shows a 1-to-1 function, a bijection, or in other words a monotonous function. If you had a periodic function for f(x) or another non-monotonous function, you would need a little bit more attention.

http://compasstech.com.au/TNSINTRO/TI-NspireCD/mystuff/showcase.html