Question

I have a fun integral.

Answer 1

\[ \int\limits \frac{\int\limits \frac{\int\limits \frac{\int\limits \frac{...}{x}dx}{x}dx}{x}dx}{x}dx\]

Repeating to infinity.

Answer 2

Strictly speaking, I assume those are different variables, like x1, x2, x3.....

Answer 3

Kinda simple, you just have to treat each integral as a sum, but that requires a lot of labor that I don't feel like doing

Answer 4

x

Answer 5

just a guess

Answer 6

^ Yep, times an arbitrary constant -- my guess for the moment.

Answer 7

log of x

Answer 8

\[\ln(\infty)\]

Answer 9

thats infinity

Answer 10

Must be a recursive function so this may have to do with squences

Answer 11

^thank you captain obvious

Answer 12

y= cx?

Answer 13

Lol

Answer 14

It's derivative is itself over x -> I gotta stick with my earlier guess of A*x, A being constant.

Answer 15

oh... I got beat lol

Answer 16

XD

Answer 17

Should really read first XD

Answer 18

Call the answer y, whatever it is.\[y=\int\limits \frac{\int\limits \frac{\int\limits \frac{...}{x}dx}{x}dx}{x}dx\] But really since it's infinite, then y is also inside the integral itself.\[y=\int\limits \frac{y}{x}dx \]
differentiate both sides \[\frac{dy}{dx}=\frac{y}{x}\]
y=Cx

Of course you could have just differentiated in the first step and got:

y=y'=y''=y'''=...

Which doesn't seem to be the same answer. Haha beats me, I just said this was an interesting integral not necessarily convergent.

Answer 19

omg how did i not think of that...

Answer 20

No, that last bit isn't true -
y' = y/x
y'' = y'/x - y/x^2 = y/x^2 - y/x^2 = 0.

Answer 21

Maybe you can use limits to awnser the integral, I know its strange but maybe using this

\[Lim((dy/dx)->infiity) \ln(x)\]

Answer 22

Oh noes... there should not be an 'outermost' integral after differentiating once D:

Answer 23

No, that's not right. The first derivative is the original thing divided by x.

Answer 24

@Jemurray3 You're right I think.

Answer 25

oops :D

Answer 26

Or no im wrong

Answer 27

Let's see, if we plug in the y into itself one step higher, do we get a different answeR?

Answer 28

lol @Matthew071
think on it... it's the same way you answer something like
\[\Large \sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}=\color{red}?\]
^_^

Answer 29

y'+xy''=y/x Is what I get.

Answer 30

Strictly speaking that's true, but as I mentioned above, y'' = 0.

Answer 31

OH MY @terenzreignz 0.0

Answer 32

How did you get to that reasoning that y'' and higher are =0?

Answer 33

Once more:
y' = y/x
y'' = y'/x - y/x^2 = y/x^2 - y/x^2 = 0

Answer 34

@terenzreignz I forgot that limits need to be plugged into x

Answer 35

Haha I'm sorry to make you copy past that @Jemurray3

Answer 36

But I believe that result can only come up after you've already chosen to plug in y to itself there. We could have: \[y=\int\limits \frac{\int\limits \frac{y}{x}dx}{x}dx\] which I feel is just as legitimate but doesn't seem to give us that same result.

Answer 37

No worries. Also, if we're being strict about notational stuff, it should be defined like this:

\[ y= \int^x \left( \frac{\int \frac{\int \frac{...}{x_3} dx_3}{x_2} dx_2}{dx_1} \right) dx_1 \]

Answer 38

Ah. See, your question is exactly *why* we need to be careful. The y on the left and the y that you've put on the right are not functions of the same variable.

Answer 39

Yeah, I was simply lazy and figured it would more or less be understood. If not, I was still interested in seeing what kind of weird things people came up with.

Answer 40

They're not the same variable, but they seem to be equally as arbitrary. The x's are all dummy variables really, so it shouldn't matter I think.

Answer 41

No no no, they *aren't* all dummy variables. The last one, on top of the last integrand, is not. That's why we can differentiate with respect to it.

Answer 42

The others are. This is a function only of x, not x1,x2,x3...

Answer 43

sorry, integral sign, not integrand.

Answer 44

I don't really know, I just sorta came up with this on my own so it was poorly defined in the first place, I didn't know this was an actual thing but what you're saying makes sense, thanks! =)

Answer 45

I guess what I'd like to know is, how would I have to define it to make it so that it did behave like I described, where I could plug it into the second one instead of the first.

Answer 46

I've never seen it before either, but it seems perfectly well-defined. You just have to be exceptionally careful about your notation. The culprit is the sloppy notation we're all taught when learning calculus:
\[ F(x) = \int f(x) dx\]
Doesn't strictly speaking make sense, and should be replaced by
\[ F(x) = \int^x f(x_1) dx_1 \]
or something equivalent. But that's a lot of extra notation and would be confusing to new students, so it's a reasonable oversight at first.

Answer 47

Just like I did would be fine, I'm not sure of the problem you have with it.

Answer 48

I suppose I don't see the problem with defining all the variables as really being like: \[y=\int\limits \frac{\int\limits \frac{\int\limits \frac{...}{x_1}dx_1}{x_1}dx_1}{x_1}dx_1\]

Answer 49

The problem with that is that the right hand side is a constant. An integral with no free variables is just a number. Not putting bounds on it doesn't make it a function, that's the sloppy (but initially acceptable) notation I mentioned earlier.

Answer 50

In which case, you're reusing the dummy variable every time on the right, which is a definite no-no.

Answer 51

Not sure you're answering my question... How is it a constant if I'm saying that y is a function of x_1?

Answer 52

\[y=\int\limits^{x_1} \frac{\int\limits^{x_1} \frac{\int\limits^{x_1} \frac{...}{x_1}dx_1}{x_1}dx_1}{x_1}dx_1\] To be more explicit.

Answer 53

I am, but let me start off more simply.
Do you recognize that this:
\[ \int_0^x f(x) dx \]
is complete nonsense?

Answer 54

I don't haha, show me the light.

Answer 55

Dont mind me Im just watching and learning :P

Answer 56

This is so intresting

Answer 57

Suppose I redefine this as just an operator that takes some quantity, divides it by x then does a linear transformation on it which is an integration. Then by doing that an infinite amount of times I converge to an answer maybe. I'm just sort of playing around, this is fun thanks @Jemurray3 and @Matthew071

Answer 58

Perhaps it would make more sense if I showed it to you in terms of a finite sum?
\[ \sum_{i = 1}^i f(x_i) \Delta x_i \]
Does that look a bit more nonsensical?

Answer 59

Sure, I guess.

Answer 60

It's like saying "I am allowing the variable x to take on values between 0 and x". It doesn't make any sense - it's referencing itself.

Answer 61

So in words,
\[ \int_0^x f(x) dx\]
Is saying
"I integrate the function f(x), allowing x to run from 0 to x"

Answer 62

It would be interesting to program this. But It would crash anybodies computer whoever tries to run it XD

Answer 63

You wouldn't really be able to - the variable is defined in terms of itself. This is a symbolic problem, and a very simple and straightforward one if you can see through the notation.

Answer 64

Sure, but the infinite integral thing we had to begin with already seemed nonsensical and self referential to begin with.

For example, if I say I just take cosine of cosine of cosine infinitely many times, I'm not actually taking the cosine of anything which seems like nonsense.

But if we say it converges to some number called x\[x=\cos(\cos(\cos(\cos(...\]inverse cosine both sides\[cos^{-1}(x)=\cos(\cos(\cos(\cos(...\]but that's just itself again. So even though we were really self referential here we were still able to get an answer and we can also cut the chain at a different place since it's infinite
\[x=\cos(\cos(x))\]
Which gives you the same answers for x. Check it on wolfram alpha, it's around .739

I guess I don't really like what definition of integral you're using, since it's not what I want. Or am I just plain out not allowed to do something like I just showed for cosine because integration is just too different from cosine?

Answer 65

Integration is vastly different. You're not just plugging a number into a function.

Answer 66

So if we have our basis as {1,x,x^2,x^3,x^4,...} then we could easily define an integration and divide by x matrix.

Answer 67

Let me try one more way to explain it - when you integrate a function of x, you need to specify the range of values you want x to take. Those are the upper and lower bounds of the integral. The upper and lower bounds are COMPLETELY independent of the dummy variable of integration. Does this make sense?

Answer 68

Forget operators and function spaces and infinite dimensional matrices for the moment. This is a very fundamental idea so it's important that it makes complete sense.

Answer 69

Yes, I understand that's what you're saying. But I'm talking about WHAT IF our bounds of integration are inseparable from our variable of integration? Sure it seems like nonsense, but I feel like it doesn't matter much since I've already said I'm doing an infinitely recursive integral in the first place which seems like nonsense to begin with.

Answer 70

No, the infintely recursive integral is not nonsense. There is not IF our integration bounds are inseparable from the dummy variable of integration - it's simply not possible. Not because I haven't ever thought of it, but because it makes about as much sense as
\[ \frac{ d^2 f}{dx^3} \]

Answer 71

Let me rewrite your initial statement of the problem in absolute, excruciating detail:

Answer 72

Sure, so does:

cos(cos(cos(cos(cos(...

There's nothing there.

Answer 73

\[ y(x) = \int^x \frac{\int^{x_1}\frac{ \int^{x_2} \frac{ ... }{x_3}dx_3}{x_2} dx_2 }{x_1} dx_1 \]

Answer 74

No, I think you're trying to make my problem into another problem it's not.

Answer 75

By saying that the sequence cos(cos(cos(cos( ... goes on forever, you are saying that y = cos(y). Don't confuse "the function is not well-defined" with "the function is defined in a confusing way"

Answer 76

And I regret to inform you that you are incorrect, and that the way I have defined the problem above is the way that you would need to write it if you were being strict. Having a recursive integration with the *same* dummy variable each time is complete nonsense, and if you speak to a local professor of mathematics then perhaps he or she can help you to understand this.

Answer 77

It is not nonsense in the sense of "undefined, infinite, incalculable" but nonsense in the sense of incorrectly applied notation.

Answer 78

y=cos(y) was a consequence of defining a mysteriously infinite function of itself without anything extra.

I don't mind being wrong, I just want to know why and I feel like I just don't see what the problem is in this case lol.

Answer 79

Another perfectly legitimate way to define your integral would be
\[ y(x) = \int^x \frac{y(x_1)}{x_1} dx_1 \]
I'm afraid I don't know how to make it more clear - you are mixing up dummy variables.

Answer 80

Perhaps this would be helpful: What is
\[ \int_0^x y \space dy \]
?

Answer 81

\[\frac{x^2}{2}\]

But what's this? \[\int\limits_0^yydy\]
Yes, you can't calculate this. But I want to give it meaning by extending it in this infinite context.

Answer 82

It is nonsense!

Answer 83

So is adding up an infinite number of things. It can't be done. But we define such things any way.

Answer 84

That's not true. An infinite number of infinitely small things can be finite - that's what integration is.

Answer 85

You are asking for the area under the curve f(y) = y from y = 0 to y = y. Do you not see how this is nonsense?

Answer 86

Yeah, I know. But you aren't doing the adding by hand.

Answer 87

Have you heard of fractional calculus? What's the 1/2th derivative mean?

If x*x*x=x^3 then what's x^(1/2)=? x multiplied by itself half of a time?!

There's nothing wrong with expanding things past their original creation.

Answer 88

You are not extending the definition of anything. I cannot help you further if you are truly not going to listen to what I'm trying to tell you.

There is a difference between extending a definition and not understanding what you're writing, and you suffer from the latter case. Let me try one final time, and then I'm going to stop, because we seem to be getting nowhere.

Answer 89

Haha it's fine, I know exactly what you're saying. It's alright to disagree.

Answer 90

You say
\[ \int_0^x y dy = x^2/2\]
and I agree with you. So I then ask, okay, what is
\[ \int_0^x \left(\int_0^x y dy \right) dy \] ?

Answer 91

Some other human defined that you know.

I'm not trying to define that. I'm trying to define:

\[\int\limits_0^x(\int\limits_0^xxdx)dx\]

Answer 92

Don't you see how this makes no sense? Once you've finished integrating with respect to y, then the variable y is GONE. The variable y on the inside is NOT THE SAME as the variable y on the outside!

Answer 93

The fundamental theorem of calculus is this:
\[ \frac{d}{dx} \int^x f(y) dy = f(x) \]
If you don't understand how this works, then you don't understand what you're doing when you integrate. If you're not willing to accept the possibility that you're trying to define something that makes no sense (it's not original, it's not new, and it's not revolutionary - it literally is a mistake in understanding that you do not recognize) then I cannot help you further, and my time on this thread must draw to a close.

Answer 94

I like your curiosity, though, so I certainly hope that you DO understand what I'm trying to say. If not now, then soon, because the longer you don't understand this point the worse it's going to get.

Answer 95

Yeah, I think you're having trouble thinking outside the box with me on this one that's all. I'm not trying to change calculus I'm just playing around by changing the rules a little.

Answer 96

Fine, let me ask you a question then.

What is
\[ \frac{d}{dx} \int_0^x x dx \]
?

Answer 97

I don't want this to come across the wrong way but this is very similar to somebody asking "What if a circle had four corners?"

You wouldn't be thinking outside the box, or expanding your idea of a circle - you'd be asking a nonsensical question. Essentially, you have to realize that a jumble of English words ending in a question mark does not necessarily form a coherent sentence -- and a jumble of mathematical symbols on a page does not form a legitimate mathematical expression.