# Integral of 0

mathelord
first the integral of 0 should be a constant rather any constant,which makes the integral of 0 to be 1=2=3=4..........
does this equate all numbers or am i gettin this mixed up.
well that is not want i intend sending,find the derivative of 3^3^3^3^3^3....................^x,and then tell me what the integral of 0 really is.I rest my case

Icebreaker
You're not making sense. The antiderivative of 0 is C, that is because information is lost when deriving. The derivative of any constant is 0. Where does "1=2=3=4" come from?

What's the difference between "a constant" and "any constant"?

Also, the definate integral of 0 on any interval is 0.

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If you have an integral of zero, and you evaluate it, you will get 0+c. C is "a constant" But you cant just pick and choose a value for c. In order to find the value of c, you need the value at a point on the function. This will determine the PARTICULAR value of c you have to use. So its not literrally "any constant", it depends on what value the curve has. 1=2=3=4 does not happen, I am not sure how you came to that conclusion, but I think it was via an error in an assumption along the way.

Alkatran
Homework Helper
Maybe it's time mathematicians start putting up massive billboards:

'MANY TO ONE' FUNCTIONS DO NOT SHOW THAT 1 = 0. 'ONE TO MANY' FUNCTIONS DO NOT SHOW THAT 1 = 0. 1 DOES NOT EQUAL 0.

HallsofIvy
Homework Helper
I think I saw one of those on I95!

mathwonk
Homework Helper
2020 Award
you guys do not seem to realize that the word "integral" does NOT mean antiderivative. the integral of zero, over any interval at all, is definitely just zero.

mathwonk said:
you guys do not seem to realize that the word "integral" does NOT mean antiderivative. the integral of zero, over any interval at all, is definitely just zero.

Yes! Exactly! And with this we are free to muse about the integral on its own terms -- it is its own thing. Which allosws us to ask the question: what is zero growing into as we add zero to it?

Such a question is a geometric-like way of thinking of the integral. And, as such, no one needs to think about antiderivatives to realize that such a thing is "definitely just zero."

What ever happened to the constant of integration?

I cant read your text because its not long enough rach, sorry. I always thought evaluating the integral was to do the antiderivative to it. Could you explain the difference please.

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quasar987
Homework Helper
Gold Member
The integral evaluated from a to b is 0, but the improper integral, i.e. the anti-derivative, is C.

rachmaninoff
Specifically, the FTC tells us that $\int_a^b 0 dx= F(b)-F(a)=C-C=0$. An antiderivative is just a function, but an integral is a number.

MalleusScientiarum
You don't even need the fundamental theorem of calculus. You can start from the Riemann-Steltjes definition of the integral and prove that any finite sum of zeros is zero.

quasar987
Homework Helper
Gold Member
Of course, but it's nice to see how the two "views" concord with each other. I.e. how the FTC "makes it work".

rachmaninoff
MalleusScientiarum said:
You don't even need the fundamental theorem of calculus. You can start from the Riemann-Steltjes definition of the integral and prove that any finite sum of zeros is zero.

Hey, I was just trying to make things easier to understand. If you're looking for the most general result, then might as well point out that a zero function (real or complex) has Lebesgue integral zero over any measurable set in any measure space.

Ah, yes of course. Thanks rachmaninoff. From the wording of the origional question, I thought it was in refrence to doing the antiderivative.

mathelord
so what does the derivative of the function 3^3^3^3^3^3^3^3^...........................^x gives us

Icebreaker
$$\frac{d}{dx} 3^x = 3^x\ln 3$$

$$\frac{d}{dx} 3^{3^x} = 3^x \ln 3 * 3^{3^x} \ln 3$$

You can do the rest. However, I don't see how this has anything to do with the integral of zero or how 1=2=3=4.

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mathelord said:
so what does the derivative of the function 3^3^3^3^3^3^3^3^...........................^x gives us
Let the function $T(a) = 3^a$. The derivative of this function with respect to a is $3^a \ln(3)$. Thus you have the function T(T(...(T(x))...)) where T is composed n times with itself. Let $3_i(x)$ be the power tower of 3 to order i where the ith position is replaced with the variable x, and $3_i$ be the power tower of order i. The derivative of the given function is then $$\frac{d}{dx}T(T(...(T(x))...)) = T'(T(...(T(x))...))*T'(...(T(x))...)*...*T'(x)$$
$$= \prod_{i=1}^n \ln(3_{i-1})*3_i(x)$$
Nasty looking thing. :yuck:

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mathelord

LeonhardEuler
Gold Member
Do you mean (3^3^3^3...^3)^x or 3^(3^(^3(....3^(x)))))))... . If you mean the first case then this is just(3^3^3^3^3...^3)^x = e^(ln((3^3^3^3^3....)^x))= e^(x*ln(3^3^3^3^3...^3)) and the derivative is therefore ln(3^3^3^3^3...)e^(ln((3^3^3^3^3....)^x)) = ln(3^3^3^3^3...)(3^3^3^3...^3)^x. In the second case it is like hypermorphism suggested.

Icebreaker
What does this have to do with your "case" that a number is equal to any other?

mathelord
no it has nothing to do with that,all i asked mainly was to find the derivative of the function.3 is raised to 3 which is raised to another 3 and so on eventually the last three is raised to x.hope u get it this time

Icebreaker
What, I didn't have it before? All you have to do is apply the chain rule.

rachmaninoff
How many 3's are there?

mathelord
infinite 3ssss
and the last three carries the x.what i mean is the latter of leonhardeuler expression.
and i do not understand wat hypermorphism did