Why is it difficult to integrate x^x

[matt grime]

In this alternate history when a student asks about the anti-derivative of 1/x he would receive the same type of response that Matt Grime gave to the original poster of this thread (as a form of parody the only thing I changed from his post was the function being given by 1/x instead of x^x).

I think you have misunderstood what I said.

Asserting that log(x) is the 'anti-derivative' of 1/x is exactly the same as declaring a symbol, F(x), I think in this case, that satisfies F'(x)=x^x.In fact, there is a well known function, erf, that is purely defined as being an integral. And very useful it is too.

Numerical integration has nothing to do with it.If you look at the preceding answers to mine, you'll notice that about 6 people all said exactly the same thing, and a complete answer it was as well. That is why there is a suggestion to lock the thread. This is also a particularly frequent discussion held in maths forums that doesn't go anywhwere.

 

[arildno]

confinement:
There are numerous functions floating about that have shown themselves useful to define in a "non-standard" way, for example diff.eq solution types of functions like the Airy, Bessel, Hankel-functions and a lot of others.

The basic reason why nobody has bothered to attach a brand new name for the anti-derivative of x^x is NOT that mathematicians are "snobbish", but because nobody has found such a frequent use of that function that it would be convenient to devise a short-hand name for it.


Find some nice use for the anti-derivative of x^x, and people will readily call it the Peterson function or what's-your-last-name-function.

 

[JANm]

Hello Arildno,
Wasn't there in this forum not also a thread where was stated that inversion of y=x^x was not possible? From there it was usefull to look at y=|x|^x at first. Perhaps it is possible to integrate |x|^x and from there one can explain that the measure of the discontinuity-set is not zero for the integral of x^x in the region x<0!
greetings Janm

 

[JANm]

This problem interested me some more.
I think mathematicians get so tired of explaining students all day long that squareroot(4)= plus or minus 2, that they forget that in the evening. For x < 0 x^x is negative if x is odd and positive if x is even, but the function is twovalued for x is even so if you take the negative value of x^x in that case then you get x^x <0 if x<0.
I therefore suggest that x^x= sgn(x)*|x|^x.
The only problem point remaining is the jump from f(x) in x=0.
That is 2 * dirac delta(x).
greetings Janm

 

[Gib Z]

I think mathematicians get so tired of explaining students all day long that squareroot(4)= plus or minus 2, that they forget that in the evening.

I don't know what kind of Mathematician would tell you that, they would stress that sqrt 4 = 2, and not -2, though both would be solutions to x^2 = 4.

 

[JANm]

I don't know what kind of Mathematician would tell you that, they would stress that sqrt 4 = 2, and not -2, though both would be solutions to x^2 = 4.

Hello Gib Z
Historically the solution to sqrt(4)=2. After some free time for mathematicians to find out that the possibility -2 is also there the democratic hazard has taken its toll. 2 seems still a better solution. It has become the principal solution to x^2=4, while -2 is a more significant solution. Yet two days ago I found a failure in my reasoning: x^2=-4 has also two solutions but those are 2i and -2i.
In comparison to x^3=-8 where the one real solution is -2. So some of my remarks in this thread are wrong x^x for x<0 is far more complicated than I made it seem in the last remarks. Sorry for my optimism...
greetings Janm

 

[maze]

I don't know what kind of Mathematician would tell you that, they would stress that sqrt 4 = 2, and not -2, though both would be solutions to x^2 = 4.

Most mathematicians I know would say that it doesn't matter what convention you use so long as you define your terms and stay consistent in your usage.

 

[Gib Z]

Most mathematicians I know would say that it doesn't matter what convention you use so long as you define your terms and stay consistent in your usage.

I was assuming the mathematician would stick to previously established conventions, and by the usual definition of the square root function, the positive value is taken.

 

[protonchain]

I can get you a definition for it I think.

From mathematica, \frac{d}{dx} x^x = x^x(1+ln(x))

So \int{x^x(1 + ln(x))} dx = x^x
\int{x^x + x^x*ln(x)} dx = x^x
\int{x^x}dx + \int{x^x*ln(x)}dx = x^x

So finally
\int{x^x}dx = x^x - \int{x^x*ln(x)}dx

Although the 2nd integral is very difficult so solve probably.

 

[JANm]

So finally
\int{x^x}dx = x^x - \int{x^x*ln(x)}dx

Although the 2nd integral is very difficult so solve probably.

Hello protonchain
For x>0 there are no problems. For x<0 there seems to be a lot of discontinuity. If x<0 and uneven x^x is real <0.
If x<0 and even then x^x is complex and has two values. Of rational numbers can be decided whether they are even or uneven, but algebraic irrationals like -sqrt(2) and trancendentals like -e or -pi one cannot decide whether they are even or uneven.
That was the problem of the integral of x^x, but since you want to know the integral of ln(x)*x^x perhaps some of this discontinity problems resolve...
good luck, Janm

 

[protonchain]

Hello protonchain
For x>0 there are no problems. For x<0 there seems to be a lot of discontinuity. If x<0 and uneven x^x is real <0.
If x<0 and even then x^x is complex and has two values. Of rational numbers can be decided whether they are even or uneven, but algebraic irrationals like -sqrt(2) and trancendentals like -e or -pi one cannot decide whether they are even or uneven.
That was the problem of the integral of x^x, but since you want to know the integral of ln(x)*x^x perhaps some of this discontinity problems resolve...
good luck, Janm

That's very true, good point. I guess the only thing left is to just hard code a definition for the integral of x^x and then treat it like an error function.

 

[JANm]

A healthy contrasting viewpoint comes from V.I. Arnold, who solved Hilbert's 13th problem, and who states that mathematics is a subfield of physics and that the endless abstractification has bogged down mathematical education. The point is that there is room for both points of view, the modern one that anti-derivatives and elementary functions are just accidental arbitrary questions with no deep mathematical structure, and the classic view, held by Euler, that the functions of interest should be expressible in terms of formulae.

Hello confinement
I am sorry to say that I don't agree with VI Arnold. Physics should be a part of mathematics. Mathematical things are correct for about 99,9 % as pure as Aluminium can be made with electrical equipement, expensive (in energy), but there is no metal alloyment which can be made so pure. So pure even that sometimes another alloyment is specifically made to improve material qualities. Rust is one: Aluminium rusts on place but in such a thin layer and so close that that is the protection. Softness, bendability etc. One thing also very interesting is it is not magnetisable so in some way it belongs to the fine metals, but it let's magnetic fields to go through. The experiment with a metal ball and a magnet under the table works with aluminium too. Apart from that it is used in the watt meter. Cannot explain right now how that works, it is called surface currents or something.
I always use the row Mathematics, Physics, Chemistry, Biology and Medics to explain how professions hang together. You are not going to say to me that mathematics is an empirical profession, are you?
That some space is needed to make the pythagoras theorem work?
Mathematicians use special addings to explain when or where theorems work, so on a flat surface a rectangular triangle etc...
that is 100 % sure. Mathematics don't give rules to anything like laws: one should not kill or steal; they give facts. The times it were laws, mathematicians always found new solutions.
Examples: you cannot take the squareroot of negative numbers: this gave the complex numbers...
A*b = b*A commutativity always holds gave the quaternions...
quaternions gave vectors and last but not least:
You cannot differentiate a step function gave the genial Dirac function. This yourney goes on. The struggle of mathematicians against formal laws that things aren't possible. As far as this struggle is at this very moment Physicians have to except it. A more pragmatic rule for physicians is the situation of technics. This is the boarder between theoretical physicists and empirical physicists. Most are of the second kind, but what I cannot stand is their attitude to their collegues who filosofically state something which cannot be measured at this time but is logically sound that that is discussioned to pieces because it cannot be measured at this time. No their diligiance goes further they state things in mathematical area which they do not own.
Personally I dislike curvature of space as empirical stement about mathematical area. In the first place phisicians tell mathematicians to calculate so difficultly that I don't know if there are many mathematicians on this world who can manage that! While if you just state that light is bended in straight space that can be calculated. Curves in space can be calculated; shapes in space can be calculated, so what have physicians to mind MATTER OVER MIND to the mathematicians? Why can't they be colleagues and fight together against the anti-technicals who are there so very much?
greetings Janm

 

[arildno]

You cannot differentiate a step function gave the genial Dirac function.

Sure you can, in a distributive sense.

 

[disregardthat]

Hello protonchain
For x>0 there are no problems. For x<0 there seems to be a lot of discontinuity. If x<0 and uneven x^x is real <0.
If x<0 and even then x^x is complex and has two values. Of rational numbers can be decided whether they are even or uneven, but algebraic irrationals like -sqrt(2) and trancendentals like -e or -pi one cannot decide whether they are even or uneven.
That was the problem of the integral of x^x, but since you want to know the integral of ln(x)*x^x perhaps some of this discontinity problems resolve...
good luck, Janm

x^x has an infinite number of values for negative and irrational x, and a finite number of values for negative and rational x. It does also have real values for all rational x=a/b where b>1 is an odd integer, a is a negative integer and the fraction is simplified. In fact, if x=a/b then x^x has b values.

 

[JANm]

I also want to comment on the snobby mathematical tone in this thread, as if many are saying 'mathematicians do not dirty our hands with such matters as finding anti-derivatives, we simply give an abstract set-theoretic definition of functions and then say that we know everything about them. Now let me return to my rigorously derived trivialities in point-set topology.' The culuture of mathematics is not owned by Hardy, Bourbaki et al but so many folks these days act like it is. A healthy contrasting viewpoint comes from V.I. Arnold, who solved Hilbert's 13th problem, and who states that mathematics is a subfield of physics and that the endless abstractification has bogged down mathematical education. The point is that there is room for both points of view, the modern one that anti-derivatives and elementary functions are just accidental arbitrary questions with no deep mathematical structure, and the classic view, held by Euler, that the functions of interest should be expressible in terms of formulae.

Hello Confinement
As mathematician even I can't make anything of the site you mentioned. I have some questions for you
1 What is the difference between a anti-derivative and a integral?
2 What is the 13th Hilbert problem?
3 How come you are so dissatisfied with tabular solutions and more rely on formula as Euler has stated?

In my historics of mathematical "learning and understanding"
Highschool: squareroot (-1) impossible, therefore: Complex numbers.
University: differentiating stepfunctions impossible, therefore distributions and the Dirac function. So there seems to be a law that when somebody says something is impossible that there come mathematicians to contradict that law. Isn't that evolution?
It seems that laws are not wholy and that nature is always busy trying to contradict these habitual thoughts of time. With mathematics it is simple 99% of the statements are proven and one learns to replicate the proof (I always controlled the proof, while learning it. When I thought I found a unlogical step in a proof I could not sleep all night and after mentioning it to the professor in charge I was set at place: A proof is a proof and is correct. So I have two questions remaining one in graph theory where of a set the mean is taken and immediately assumed that this is an element of the set. With an uneven number of elements that is correct but with an even number of elements there is a choice problem. Secondly with the proof of innumarability of numbers there is used suppose a numbrification we construct a number that does not fit any numbrification so innumerability is proven... these two gave me doubt, all the others not...).
Physics is different. The physical part one tries to understand with dimensions and so and the mathematical part is practiced, as far as one is capable at that moment (memory, openness to used parameters, insight), something more variable as one should think, which prooves that people are not computers. But my confrontations with physics professors are not so much different with those with the mathematics-professors. Questions are never answered as if they could (possibly) open a new insight. Real questions are only answered as boresome... Not standard answerable and agitating the docent, with probably his own problems could be, but never: at this point you give at least the impression of attending.

Im sorry Mathematicians and physicist, I was serious at your matters for more than 40 years but I am sick and tired of the closed answeres I got in al those years. Keep believing in Einstein And wikipedia and science will evolve...

 

[CRGreathouse]

In my historics of mathematical "learning and understanding"
Highschool: squareroot (-1) impossible, therefore: Complex numbers.
University: differentiating stepfunctions impossible, therefore distributions and the Dirac function. So there seems to be a law that when somebody says something is impossible that there come mathematicians to contradict that law. Isn't that evolution?

The trouble here isn't the math but the translation between math and the vernacular. Translating "there does not exist an x in R such that x^2 = -1" as "squareroot (-1) impossible" makes the complex numbers seem to break the rule, but they don't. That's why the interpretation of mathematics can be difficult and important. A good example, in my view at least, is Arrow's theorem, often vernacularized as 'there are no good voting systems' (or, worse, as 'the only good voting system is a dictatorship'). When you say it that way it sounds pretty bad!

 

[Дьявол]

\int{x^xdx}

t=x^x

dt=x*x^{x-1}dx

dt=x^xdx

dx=\frac{dt}{x^x}

\int{x^x * \frac{dt}{x^x}}=\int{dt}=
=t=x^x

I don't find it difficult to solve at all.

 

[Gib Z]

You evaluated dt incorrectly - the power rule only applies when the exponent of x is a constant - in this case it is not. To evaluate its derivative you must first convert it to a exponential form:

x^x = e^{x\ln x}

Then use the chain and product rules.

 

[0xDEADBEEF]

t=x^x

dt=x*x^{x-1}dx

Bollux. Such a mathematical rule does not exist.

 

[arildno]

You evaluated dt incorrectly - the power rule only applies when the exponent of x is a constant - in this case it is not. To evaluate its derivative you must first convert it to a exponential form:

x^x = e^{x\ln x}

Then use the chain and product rules.

Alternatively, we may differentiate it as the follows:

First, we regard the x in the base as our variable, the exponent as being constant, getting x*x^{-1}=x^{x} as our result.

Then, we let the x in the base be treated as a constant, regarding the exponent-x as our variable, getting x^{x}\ln(x)

Finally, we add the two results together, getting:
\frac{d}{dx}x^{x}=x^{x}(1+\ln(x))[/itex]<br /> which is, of course, the right answer..<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" />

 

[Gib Z]

I can see a vague intuitive reasoning to that method but not a rigorous one, would you please provide one ?

 

[arildno]

I can see a vague intuitive reasoning to that method but not a rigorous one, would you please provide one ?

Set:
f(y,z)=y^{z}, Y(x)=x, Z(x)=x
Thus, defining
g(x)=f(Y(x),Z(x))
we get:
\frac{dg}{dx}=\frac{\partial{f}}{\partial{y}}\frac{dY}{dx}+\frac{\partial{f}}{\partial{z}}\frac{dZ}{dx}=\frac{\partial{f}}{\partial{y}}+\frac{\partial{f}}{\partial{z}}

which is what we were after.


(note that this amounts to a generalization of the product rule; if h=f*g, the h'=f'*g+f*g', i.e, we may differentiate the factors separately, and then add the results)

 

[zcd]

Can't \frac{d}{dx}x^{x} be done by implicit differentiation?

y=x^{x}

\ln(y)=x\ln(x)

\frac{1}{y}\frac{dy}{dx}=\ln(x)+1

\frac{dy}{dx}=y(\ln(x)+1)=x^{x}(\ln(x)+1)

One of the finer moments in first semester calculus ab

 

[arildno]

Can't \frac{d}{dx}x^{x} be done by implicit differentiation?

y=x^{x}

\ln(y)=x\ln(x)

\frac{1}{y}\frac{dy}{dx}=\ln(x)+1

\frac{dy}{dx}=y(\ln(x)+1)=x^{x}(\ln(x)+1)

It sure can, there are many ways to do this.

 

[Elucidus]

Greetings all.

Some comments on previous posts.

Original question concerning the integral of x^x:

Although x^x being continuous for all real x > 0 implies it has a continuous antiderivative, the function (which exists) has not yet been labeled. It has been shown to be non-elementary (meaning it is not an algebraic combination of polynomials, radicals, trigonometric, logarithmic, or exponential functions or their inverses). Mathematica and the like will spit back the unevaluated integral since the function is not explicitly detailed in its library of functions (elementary or no).

I believe it has been shown that no proper subset of the continuous functions on R is closed under integration. So that no matter what set of functions you start with (unless you have all of them) one can construct an algebraic combination of them whose antiderivative is not a member of the set.

Second comment: x^x is defined for only a sparse number of values less than 0. Even if you permit extension to complex values or sheets, the set upon which some sort of defintion is possible has Lebesgue measure 0. Integrating over such a set is rather fruitless. Classically, analysis of the function x^x is restricted to the sensible domain of x > 0.

Side note: Euler was able to calculate the area under the curve y = x^x between 0 and 1, but I have misplaced my reference to the proof.

Math and Physics:

The debate about the relation of Mathematics and Physics (as well as Mathematicians and Physicists) is old and overwrought. Knowing Physicists and being a Mathematician I can assure you that neither is a subdiscipline of the other. One resides in an ideal universe of abstractions while the other lives and explores the natural universe. Fortunately for humanity much can be learned from each other. The area of Mathematical Physics (from whence one of the Millenium probelms arises) is rich and deep, but there are areas whithin each discipline where the other just fails to be relevant. Any claim that one is included in the other is demonstrably false.

Breaking Mathematical Laws:

Saying that sqrt(-1) is undefined is broken by the development of the complex numbers is bit of a misstatement. The function sqrt() has different meanings in different contexts. sqrt(x) as a real-valued function of a real variable is defined only for x >= 0. It is technically called the principal square root function (and is also a unary operator on the nonnegative reals). Even though (-3)^2 = 9, the sqrt(9) = 3, since the function must be single-valued. The sqrt(-1) is undefined.

When mathematicians tried to answer the question "what if we made a number whose square is -1, what would happen?" the function sqrt() was changed to mean something else. In this new environment, sqrt(-1) was defined to be the imaginary unit i. Unfortunatley it changed the meaning of the root to something other than the principal root. By extension of DeMoivre's Theorem, if you represent a complex number in trigonomtric form (r, t) where r is the complex modulus and t is the complex argument (in radians) one can define the primitive nth root to be (r^(1/n), t/n). Unfortunately there is discrepency between the (real) principal cube root and the (complex) primitive cube root. cbrt(-8) = -2 in the real sense but cbrt(-8) = 1 + i.sqrt(3) in the complex sense. -2 is the second order primitive cube root of -8 (i.e. cbrt(-8)^2).

One must be clear about the context of the functions. Shifting context from one environment to another willy-nilly can get many into trouble (as my students have often demonstrated). Make sure one is comparing apples with apples.

Enough for now.

Thanks for your time.

--Elucidus

 

[Дьявол]

Thanks for the correction everybody.

Here is my new solution.

If
<br /> x^x = e^{x\ln x}<br />
then

\int{x^xdx}=\int{e^{x\ln x}dx}

Now using the substitution method:

t=e^{x\ln x}

and

dt=e^{x\ln x} * (1+\ln x) dx

Now

dx=\frac{dt}{e^{x\ln x} * (1+\ln x)}

and substituting in the original equation:

\int{e^{x\ln x}dx}=\int{t*\frac{dt}{t * (1+\ln x)}}=\int{\frac{dt}{1+<br /> \ln x } }

Is this better? But somehow I just need to present ln(x) in terms of t.

 

[arildno]

Quite so!

 

[Дьявол]

Quite so!

Thanks, but it seems like there is not way presenting ln (x) in terms of t, since ln(x)=ln(t)/x, and there is x again.

 

[arildno]

Thanks, but it seems like there is not way presenting ln (x) in terms of t, since ln(x)=ln(t)/x, and there is x again.

Quite so.

You might as well give up.

 

[Pinu7]

Is this better? But somehow I just need to present ln(x) in terms of t.

Only possible if we can define and inverse function of x^x(which will only work for x>0).

And then, I doubt the integral COULD be solved analytically. However, maybe it would be simpler to approximate?

 

[arildno]

Only possible if we can define and inverse function of x^x(which will only work for x>0).

And then, I doubt the integral COULD be solved analytically. However, maybe it would be simpler to approximate?

As it happens, you will need to define two distinct inverses, one for 0<x<1/e, and one for 1/e<x

 

[Elucidus]

Side note: Euler was able to calculate the area under the curve y = x^x between 0 and 1, but I have misplaced my reference to the proof.

--Elucidus

Whoops. The integral of x^x between 0 and 1 was calculated by Johann Bernoulli in 1697 using power series (not Euler). The proof appears in "Opera Omnia" vol. 3 (1697) pp. 376 - 381.

He proved that the definite integral from 0 to 1 is Sum[n = 1 to infinity](-1)^(n+1)/(n^n) which equals (to 10 decimal places) 0.7834305107. I highly suspect that this number is not only irrational but also transcedental.

It is unlikely that his methods can be extrapolated to handle many other boundary values other than 1 or certain powers of e.

Bernoulli's proof is discussed in the book "The Calculus Gallery" by William Dunham (2005, Princeton University Press) on pp. 48 - 51.

--Elucidus

 

[Elucidus]

Thanks for the correction everybody.

Here is my new solution.

Is this better? But somehow I just need to present ln(x) in terms of t.

The function f(x) = x^x is invertible for x >= 1/e. There is a quasi-elementary function known as the Lambert W function, which I will just abbreviate as W(x), defined so that y = W(x) iff x = y * e^y. Using this one can define the inverse of x^x as:

y = ln(x)/W(ln(x)).

In the integral \int{{x^x}dx} using the substitution t = x^x, one gets dx = dt/(t * (lnx + 1)). The integral becomes:

<br /> \int{\frac{dt}{\ln x + 1}}<br />

using the inverse function mentioned earleir, ln(x) equals W(ln(t)) and the integral is:

<br /> \int{\frac{dt}{W(\ln t) + 1}}<br />

However, after wrestling with that integral for over an hour, I see no further manipulations that make it easier to work with. Power series is an option, but at that point it would have been easier to use power series from the outset.

--Elucidus

 

[Gib Z]

Whoops. The integral of x^x between 0 and 1 was calculated by Johann Bernoulli in 1697 using power series (not Euler). The proof appears in "Opera Omnia" vol. 3 (1697) pp. 376 - 381.

He proved that the definite integral from 0 to 1 is Sum[n = 1 to infinity](-1)^(n+1)/(n^n) which equals (to 10 decimal places) 0.7834305107. I highly suspect that this number is not only irrational but also transcedental.

It is unlikely that his methods can be extrapolated to handle many other boundary values other than 1 or certain powers of e.

Bernoulli's proof is discussed in the book "The Calculus Gallery" by William Dunham (2005, Princeton University Press) on pp. 48 - 51.

--Elucidus

Due to its similarity to the freshman's dream ( (x+y)^n = x^n + y^n) this identity is called the Sophomore's dream, but is actually true. Good information is found on it at :

http://en.wikipedia.org/wiki/Sophomore's_dream

http://mathworld.wolfram.com/SophomoresDream.html

 

[CRGreathouse]

The integral of x^x between 0 and 1 was calculated by Johann Bernoulli in 1697 using power series (not Euler). The proof appears in "Opera Omnia" vol. 3 (1697) pp. 376 - 381.

Oh, darn. I seem to have misplaced my copy of volume 3.

 

[JANm]

The trouble here isn't the math but the translation between math and the vernacular. Translating "there does not exist an x in R such that x^2 = -1" as "squareroot (-1) impossible" makes the complex numbers seem to break the rule, but they don't. That's why the interpretation of mathematics can be difficult and important. A good example, in my view at least, is Arrow's theorem, often vernacularized as 'there are no good voting systems' (or, worse, as 'the only good voting system is a dictatorship'). When you say it that way it sounds pretty bad!

Hello CRGreathouse
It is nice you want to complicate democratic politics to this matter.
It is not the question wether +i or -i is the solution of x^2=-1, nor a voting problem to it. They fit both. The tramp sitting on the throne or the yuppie. Both satisfie the eqaution. Since Napoleon we have had kings and your socalled dictators (in some sort of way selected bests of the bad). If they are inhuman presidents it takes four or in some countries six years to get rid of them...
Investors think they invest on the best revenue. They should know better: they should invest on what are the Better idea's. But the improvement they leave to Subsidicement of the state and filosofers who want to do that have to proove themselves in front of telivision first. Clowning your career. When dead and clown seems to have been something; the investors who bought requisites;... are rich. Bought yourself requisites of somebody nothing, alah.
Waht a stupid thing is kapitalism...

 

[JANm]

Am very sorry to have stopped this thread seemingly with all kind of political items. The only political Idea of me (as an exact scientist (try to be)) is that it would be nice if we would work with meters seconds grams or kilograms Coulombs Amperes Joules Newtons Watts etc. for it seems so very much clear that relating measurable things together fit the best that way. After that politics is me worst, sorry sausage.
The matter of integrating x^x is after that again a very interseting matter. For positive x there must be an univocal solution. For negative x there is A: the value of -1<x<0 where there is some sort of powering with the factor 1/x, while B: x<-1 is some sort of rooting.
I think it would be nice if someone explains if hier or her problem lies in the A region or in the B region. Just to concentrate on something visiulisable with graphs which are commonly known. Concentration is a thing which did not become easier with internet I must say...
greetings Janm

 

[Elucidus]

The matter of integrating x^x is after that again a very interseting matter. For positive x there must be an univocal solution. For negative x there is A: the value of -1<x<0 where there is some sort of powering with the factor 1/x, while B: x<-1 is some sort of rooting.
I think it would be nice if someone explains if hier or her problem lies in the A region or in the B region. Just to concentrate on something visiulisable with graphs which are commonly known. Concentration is a thing which did not become easier with internet I must say...
greetings Janm

The expression x^x can be defined for x < 0 only on a set of Lebesgue measure 0 (as I think was mentioned). So any integral over this region is going to be 0. The function can be made continuous for all x \geq 0 by defining it to be 1 at x = 0 and e^{x \ln x} for all x > 0. So it is Riemann integrable over [0, infinity). Unfortunately, not nicely.

--Elucidus

 

[JANm]

The expression x^x can be defined for x < 0 only on a set of Lebesgue measure 0 (as I think was mentioned). So any integral over this region is going to be 0. The function can be made continuous for all x \geq 0 by defining it to be 1 at x = 0 and e^{x \ln x} for all x > 0. So it is Riemann integrable over [0, infinity). Unfortunately, not nicely.

--Elucidus

Hello Elucidus
It is going somewhat fast to me. In the first place: defining x^x=1 for x = 0. That is put nicely because you know that even x^0 has only one possible singularity and of course that is at x=0. Let us suppose that your definition 0^0 =1 is correct, it is at least the most expected one; you state x^x=e(x*ln(x)) for x > 0 so it is Riemann integrable.
What does that mean? and why isn't it nicely Riemann integrable?

For the values x < 0, of course not |x|<0 because that is impossible the function has defined value for x included in the odd rational numbers and is two valued for the even rational numbers. Is there not a way to know of irrational numbers whether they a odd or even?
greetings Janm

 

[Elucidus]

Hello Elucidus
It is going somewhat fast to me. In the first place: defining x^x=1 for x = 0. That is put nicely because you know that even x^0 has only one possible singularity and of course that is at x=0. Let us suppose that your definition 0^0 =1 is correct, it is at least the most expected one; you state x^x=e(x*ln(x)) for x > 0 so it is Riemann integrable.
What does that mean? and why isn't it nicely Riemann integrable?

For the values x < 0, of course not |x|<0 because that is impossible the function has defined value for x included in the odd rational numbers and is two valued for the even rational numbers. Is there not a way to know of irrational numbers whether they a odd or even?
greetings Janm

Since 0^0 is not formally defined (in most places) the function x^x is continuous only for x > 0. But

\lim_{x \rightarrow 0^+} x^x = 1.

So there exists an extension of x^x that is right continuous at 0, where it is defined to be 1 when x = 0. i.e. the function

f(x) = \left\{ \begin{array}{rl}<br /> x^x, &amp; \text{if } x&gt;0 \\<br /> 1, &amp; \text{if } x=0 \end{array}

is continuous on [0, \infty).

The Fundamental Theorem of Calculus states that any function continuous on an interval is Riemann integrable (i.e. has a continuous antiderivative). Since x^x (and the aforementioned extension) are continuous then they are Riemann integrable. However, not every integrable function has a nice antiderivative. The antiderivative of x^x is known to be non-elementary and has no nice closed representation despite knowing that it must exist. This is what I meant by "not nicely" - it exists, but we can't describe it well.

Trying to extend x^x to negative values works only for a limited set of values. The function becomes highly discontinuous. Generalizing integration to the Lebesgue integral doesn't help either. Or you have to twist it into some serious complex valued contortions. Neither seems worth it.

--Elucidus

 

[John Creighto]

Thanks for the correction everybody.

Here is my new solution.

If
<br /> x^x = e^{x\ln x}<br />
then

As a side for some reason I was thinking about a good series representation.

If I expand ln(x) first then one gets

x^x=\Pi_{n=0}^{\infty}exp \left( \frac{(-x)^n}{(n+1)!} \right)

Consider what happens now if you plug in \frac{(-x)^n}{(n+1)!} for the Taylor series of an exponential. And then multiply each of these terms together. If you are only interested in the first N terms of the Taylor series, then you only need to consider the first N terms in the product.

Also looking at the series, it kind of looks like the logarithm of x^x is \frac{1}{x}exp(-x)

but I think I must have made a mistake.

 

[Dragonfall]

Guys, this thread was from 2005. Is this normal?

 

[JANm]

Only possible if we can define and inverse function of x^x(which will only work for x>0).

And then, I doubt the integral COULD be solved analytically. However, maybe it would be simpler to approximate?

Hallo Pinu7
It seems we are answering a question outdated. Do you matter?, I don't. Mathematical problems survived wars, so why not years? Ok how to integrate x^x?
You say you can inverse the function for x>0. What will that be?
greetings Janm.

 

[ansh112]

Well the best I managed is an infinite series, which curiously when evaluated from 0 to 1 gives
[[math]]
1-\frac{1}{2^{2}}+\frac{1}{3^{3}}-\frac{1}{4^{4}}+...
[[math]]
I wrote x^x as e^xlnx and used the infinite series e^x. Integrating term by term and used l'Hospital's rule and recursive nature of the integrals to generate the sequence above. Checked using definite integral calculator

 

[JJacquelin]

Hello !

just a piece of information to be added to this babylonian topic:
http://www.scribd.com/people/documents/10794575-jjacquelin
Then, select the article "Sophomore's Dream Function"

 

[powerovergame]

Hello !

just a piece of information to be added to this babylonian topic:
http://www.scribd.com/people/documents/10794575-jjacquelin
Then, select the article "Sophomore's Dream Function"

Haha that is some excellent material, pretty much answers everything here.

 

[Alejandroman8]

how integrate x^x^x ?my teacher ask me)what do you think about it?

 

[Elucidus]

how integrate x^x^x ?my teacher sak me)what do you think about it?

I think you're in for a lot of swearing and bloody knuckles. Since x^x is positive for all x > 0 then x^{(x^x)} is continuous for x > 0 and therefore Riemann integrable. But trying to find any sort of friendly resolution to it is a fool's errand.

Numeric approximation is your best hope since I think even power series will be intractable.

--Elucidus

 

[powerovergame]

how integrate x^x^x ?my teacher ask me)what do you think about it?

Let's not think about how to integrate, just think about the easy case, differentiation, you still wouldn't get nice results at all.

http://the-genius-group-from-uc-berkeley.googlegroups.com/web/Tetration%20Differentiation.pdf?gsc=JivL3wsAAABzdSJOyPIJbvk4ERCPyg5o

 

[michaelc187]

Basically, I say (cause and effect stuff/ quark stuff / physically imposside to divide stuff / blah b

going along with everything in nature can be defined as and affacting the reality as (itself is) ,,, call it single definable existence or basically the law of of cause and affect exists...: blah blah blah, no fractions or partial exitstance. (avoid s domain)

(according to the equation derivative guy, (tesla I like)... nueton (sp)...) slope!

der x to x is
is
(xtox - ((xex) - (xe1)to(xe1)) / 1

rise over run in a a non fraxctional (cause and effect world) tadah...

Take a microprossessors class in Electrical Engineering (learn machines do math etc... O'Mally Universiy of Fl... nothing better, I've seen) ...followed by a deep physicall class in statistics (math side, if it exists or maybe not (math dept, might kick your but in statistics) do stilll calll it "counting statistics? get into those sum equations/lin equations and don't get they are the same,
take a math department lin equations classes (at the same time)

that semester will be fun...