# integral of x^x

Discussion in 'Physics & Math' started by ddovala, Jun 27, 2004.

1. ### ddovalaPi is exactly 3Registered Senior Member

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How do you find it? Or at least estimate it? Any help would be appreciated.

3. ### ddovalaPi is exactly 3Registered Senior Member

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Oh, and how do you do exp(x^2) too?

5. ### CrispGone 4everRegistered Senior Member

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For x^x, try writing it as a logarithm, e.g. x^x = e^(x log x).

7. ### James RJust this guy, you know?Staff Member

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There is no closed-form expression for the integral of exp(x^2). However, the values of some particular definite integrals involving that function can be calculated, and the integral is tabulated numerically.

8. ### DunnoyetRegistered Senior Member

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I recently ran over x^x... It is one of THE wierdest things I've encountered, despite studying fractals on the side.

x^x for x in (infinity, 0) would probably be best analyzed with complex numbers. Try graphing it sometime; it comes out really wierd, even if you do pick the right values to graph. Perhaps integrating it requires the use of complex numbers... I really do want to know. The Mathematica online integrator seems to think of x^x as non-integrable.

Crisp,
I really have no idea how to integrate exponential functions. Could you expand on that? Did you actually get an answer?

9. ### CrispGone 4everRegistered Senior Member

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I didn't try anything when I posted the hint; usually the approach for avoiding problems with functions like x^x is to rewrite them as exponentials, but after having thought it over, I don't think that it will help here. You can rewrite it to an integral where you would seem to need to primitive function of ln(x), which does not exist.

Maple also seems to indicate that there is no closed form for the integral of x^x.

10. ### DunnoyetRegistered Senior Member

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Thanks! That is more of an answer than I've gotten before.

I wonder if any graduate-level math professors or students could shed some light on this oddball function.

Considering x^x where x<0, I think that truly graphing it becomes a job for complex numbers. Perhaps analysis from a heavy complex math standpoint would help...maybe?

Is there a special way to integrate functions of real domain and complex range?

Could Riemann Sums be used to estimate the integral of a positive range of x^x?

(I won't be back to reply until the seventh of July; in the meantime I'll be subscribed to the thread.)

11. ### letheRegistered Senior Member

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this function does not have an elementary integral, just as e^(x^2) and sin(x)/x do not.

This is a standard result, seen in any reference on elementary antiderivatives.

if you want to estimate it, Riemann sums should work fine, or perhaps Simpsons method.

This is a standard method of numerical integration, taught to high school calc students.

12. ### DunnoyetRegistered Senior Member

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(ok, so I'm still on...)

Ah... I will be taking Calculus this coming schoolyear, so I should learn it then...

That should answer one of ddovala's questions.

13. ### Mr. ChipsBannedBanned

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I am paying close attention to this thread. That particular function can be used to describe a specific kind of self-similar fractal that could be represented by a digraph, directed probability graph. Say 2 squared, four taken in two groups each selecting one. Each of those associate in the next "node" of the network diagram again in a group of two choosing one. The number of steps in this "pyramid" is two. Try 3, 27 in groups of three, 9 groups of three each selecting one leaving 9 for 3 groups of three leading to one group of three, again 3 steps, same for any other number. Think of a chain letter except instead think of chain conferencing. The largest number so far that would not exceed the population of humanity is 9 raised to its own value.

Anyways, appears to me, and I certainly can be mistaken as its been a while since I tried to figure it out (got lots of papers in my files that maybe I should dig up), that Crisp's approach approaches utility though I think you need to use the natural logarithm, that is, x^x = e^(x ln x) leads to the integral of x^x being e^(x ln x) + C

14. ### shmoeRegistred UserRegistered Senior Member

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This won't work. The derivative of e^f(x) is (e^f(x))*f'(x), there's that pesky chain rule term. Try to differentiate e^(x ln x) + C using this, you won't get x^x.

Dunnoyet, if you want to integrate a function with Real domain and complex range, you can just integrate Re f(x) and Im f(x) over whatever interval you like. Each is a real valued function, combine to get the complex valued integral. If you are are looking at (-x)^(-x) for positive values of x, you'll essentially have
1/(x^x) times an oscillating term (made up of a real cos term and an imaginary sin term). This can be handled numerically over given finite intervals, but that's about it.

15. ### DunnoyetRegistered Senior Member

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Okay...wow...cool...thanks!

Complex math is cool...

P.s. Is there a name for functions related to x^x?

16. ### Mr. ChipsBannedBanned

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x^x = e^(ln x^x) = e^(x ln x)

Correct?

My understanding is that the integration of e^x is e^x + C according to what I reference on the web such as at this site: http://www.gomath.com/htdocs/ToGoSheet/Calculus/integrals/integrals.html, seventh formula from the top.

That makes the integral as shared before, e^(x ln x) + C

Differentiating e^u + c results in e^u as detailed at this site, http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Exponential_Function.htm One does drop the constant, right?

I find that last site to be quite nice at explaining euler's number and will spend some more time with it. The binomial expansion definition of the term relates it right into the realm of probability distributions. I believe it may be possible to design a communication system where the probability of generating symbolic representation tends towards an ideal of greatest possible information carrying capacity, which is besides the point of this thread.

Edited to change variable to match the used reference.

Last edited: Jun 30, 2004
17. ### AndersHermanssonRegistered Senior Member

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Mr. Chips

In your case u denotes a variable, not a function. Read Schmoes post again!

18. ### Mr. ChipsBannedBanned

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Please, someone, reference me to this datum shared by Schmoe. So far I find reference to otherwise.

Functions explicitly involving euler's number seem to have the unique distinction of resulting in the same equation whether through differentiation or integration except for the added constant.

Am I entirely off my rocker here? Seems the integration of x^x is x^x + c .

19. ### shmoeRegistred UserRegistered Senior Member

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Again, you are missing the chain rule term.

The derivative of e^x wiith respect to x is e^x.

If you have a more diabolical function of x in the exponent, say u(x), then the derivative of e^(u(x)) is (e^(u(x)))*u'(x), where u'(x) denotes the derivative of u(x) with respect to x.

Let's differentiate x^x+c=e^(xlogx)+c with respect to x. Here u(x)=xlogx. We get:

(e^(xlogx))*(logx+1)+0

the 0 is from the constant term, logx+1 is the derivative of xlogx (product rule)

So the derivative of x^x+c is (x^x)*(logx + 1), which is not at all the same function as x^x.

You should be able to find a reference to the chain rule just about anywhere.

20. ### CrispGone 4everRegistered Senior Member

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As said before, you need to apply the chainrule. It is probably best understood as follows:

You know that d(e<sup>x</sup>)/dx = e<sup>x</sup>. Now suppose that you want to derive e<sup>y</sup> to x, then what you would do is:

d(e<sup>y</sup>)/dx = (d(e<sup>y</sup>)/dy) * (dy / dx)

which is just multiplying with dy / dy. Since deriving an exponential to the variable in the exponent is simply back that function, the last becomes:

d(e<sup>y</sup>)/dx = (d(e<sup>y</sup>)/dy) * (dy / dx) = e<sup>y</sup> * (dy / dx)

Now if you replace y = f(x) , y as some function of x, then you get:

d(e<sup>f(x)</sup>)/dx = e<sup>f(x)</sup> f'(x)

where I wrote f'(x) = df(x)/dx.

When "deriving to other variables than x", you always need to do the trick above. This "trick" is called "the chain rule" and it is formally written as Shmoe explained:

d f(g(x)) = f'(g(x)) * g'(x) * dx

It is called the chain rule because, if you study the right hand side above, you see that if you want to derive f as a function of g as a function of x, you get the derivative of f followed by the derivative of g followed by the "derivative" of x (a chain of derivatives). You can immediatelly generalise this to longer chains of functions.

I've been very sloppy with the language here, but I hope it somehow summarizes how you find that x^x derived is not simply x^x.

Bye!

Crisp

21. ### Mr. ChipsBannedBanned

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Cool, thanks, makes sense to me along with the references I've seen elsewhere on the web. Okay, looks like we have the derivative of x^x as (x^x)*(ln x + 1).

So, working togethor here, we ought to be able to figure out the integral, specifically, integral of x^xor e^(x ln x) in respect to x?

22. ### shmoeRegistred UserRegistered Senior Member

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Nope-take a look at lete's post. There's no closed form solution to the integral of x^x. Essentially this means the antiderivative of x^x cannot be expressed in terms of a finite number of the usual functions cos, sin, exp, polynomials, etc. This isn't just saying someone hasn't found such a closed form yet, it means it doesn't even exist to be found.

This doesn't mean an antiderivative doesn't exist, just that it isn't "nice". You can also still find a definite integral of x^x over a given interval to any degree of accuracy you like.

23. ### DunnoyetRegistered Senior Member

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I think that to integrate something involving x^x you have to have it "well formed" (I think that that is the term), in other words you have to wangle x^x into (x^x)*(ln x+1).

I wonder if any fractals show up when the mapping x => x^x is played with...