Why is it difficult to integrate x^x

[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:
[tex]\frac{d}{dx}x^{x}=x^{x}(1+\ln(x))[/itex]
which is, of course, the right answer..

 

[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
$$x^x = e^{x\ln x}$$
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:

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

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:

$$\int{\frac{dt}{\ln x + 1}}$$

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

$$\int{\frac{dt}{W(\ln t) + 1}}$$

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

$$f(x) = \left\{ \begin{array}{rl} x^x, & \text{if } x>0 \\ 1, & \text{if } x=0 \end{array}$$

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
$$x^x = e^{x\ln x}$$
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
[\(\displaystyle ]
1-\frac{1}{2^{2}}+\frac{1}{3^{3}}-\frac{1}{4^{4}}+...
[\(\displaystyle ]
I wrote x^x as e^xlnx and used the infinite series e^x. Integrating term by term and used l'Hospital's rule and recursive nature of the integrals to generate the sequence above. Checked using definite integral calculator\)\)

 

[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...