## Integral of x^x

For the discussion of math. Duh.

Moderator: gmalivuk

### Integral of x^x

As some of you may know, xx does not have an analytical integral. You can try to do it by hand, or you can get a computer to do it, but you won't get it.

I've always been curious about the xx function. After finding that the LambertW could give me its inverse, I started wondering about its integral. Although I can't get it in f(x) form, I can do the next best thing, which is graph it (first plot, compared with x^x). It diverges at a slower rate than xx, which is to be expected when you look at its derivatives.

Since dnx/dxn of xx is xx is xx(1+ln(x))n, I figured that the integral would similar to the function DIVIDED by 1+ln(x), which is a reasonable guess at high x, but falls apart for low x. The second graph is the ratio of the actual integral to my guess.

Anyway, we can use this thread for talking about one of my favourite functions, x^x.

Klotz
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### Re: Integral of x^x

You can express the integral with an infinite series:

http://en.wikipedia.org/wiki/Sophomore%27s_dream

EDIT: Unless you wanted an indefinite integral...
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mathmagic
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### Re: Integral of x^x

Oh yeah I remember that. Do you know why it's called the sophomore's dream?

I'm more curious about an indefinite integral.
Klotz
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### Re: Integral of x^x

http://forums.xkcd.com/viewtopic.php?f=3&t=2543

I think I did the integrals of

xax
xx^a

Where a is an arbitrary constant.
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### Re: Integral of x^x

I found this from a google search as to why it's called "the sophomore's dream": http://answers.yahoo.com/question/index?qid=20080411192254AAHCWVb

The wiki article really should have something about the etymology, although that would require a (real) source that I don't have.

Edit: Looks like someone added to the wiki page since I looked at it this morning. Never mind then.
auteur52
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### Re: Integral of x^x

auteur52 wrote:Edit: Looks like someone added to the wiki page since I looked at it this morning. Never mind then.

Looks like it was some guy named "skeptical scientist". Hmmm.....
skeptical - adj. open minded; willing to be convinced or dissuaded by evidence

"Any sufficiently advanced incompetence is indistinguishable from malice." - Clark's Law

skeptical scientist
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### Re: Integral of x^x

On another note, I worked out this formula for the derivative of a tetration:
d/dx nx=nx*[(d/dx n-1x)ln(x)+(n-1x)/x]
where d/dx 1x=x^x(1+ln(x)).
Klotz
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### Re: Integral of x^x

skeptical scientist wrote:
auteur52 wrote:Edit: Looks like someone added to the wiki page since I looked at it this morning. Never mind then.

Looks like it was some guy named "skeptical scientist". Hmmm.....

Conspiracy!
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Bassoon
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### Re: Integral of x^x

I feel like I have contributed to society

SS did you at all look at my thing at writing up the wiki page?
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### Re: Integral of x^x

I didn't write the page on sophomore's dream, I just added the material on the source of the name which I cribbed from the yahoo answers page.
skeptical - adj. open minded; willing to be convinced or dissuaded by evidence

"Any sufficiently advanced incompetence is indistinguishable from malice." - Clark's Law

skeptical scientist
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### Re: Integral of x^x

Klotz wrote:dnx/dxn of xx is xx is xx(1+ln(x))n

plugging in x = 1, we get that all the derivatives of xx are 1. This seems... wrong.
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notzeb
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### Re: Integral of x^x

notzeb wrote:
Klotz wrote:dnx/dxn of xx is xx is xx(1+ln(x))n

plugging in x = 1, we get that all the derivatives of xx are 1. This seems... wrong.

I get as the second derivative:
xx (lnx + 1)2 + xx-1
It seems Klotz forgot the second term when differentiating a product.
jaap
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### Re: Integral of x^x

I feel slightly less accomplished now.

I want to make some contribution to math, somehow, soon hopefully
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### Re: Integral of x^x

Okay, this isn't completed work, but I thought I'd flag the idea in the hope that someone else might know what the upshot will be. People are saying this indefinite integral cannot be expressed in terms of elementary functions. Fair enough, but what if we throw in the Lambert W, what happens then? Can we do it then?

My progress so far, and really this needs other eyes on it to see if I've made a hideous blunder. We start by making a substitution x^x=z which WP tells me is the same as letting
x=\frac{ln(z)}{W(ln(z))}=exp(W(ln(z)))
which, using the later expression, gives us
\begin{align*} \frac{dx}{dz} &= \frac{d}{dz}[exp(W(ln(z)))]\\
&= exp(W(ln(z))) \frac{d}{dz}[W(ln(z))]\\
&= \frac{ln(z)}{W(ln(z))}\cdot \frac{W(ln(z))}{ln(z)(1+W(ln(z)))} \cdot \frac{1}{z}\\
&= \frac{1}{z(1+W(ln(z)))}\end{align*}

which allows us to say that
\int x^x dx = \int \frac{dz}{1+W(ln(z))}.

Now, this isn't too hideous, especially if you let t=ln(z), but maybe that's a serious misstep. Anyways, any thoughts? Is this heading towards a brickwall or not?

Edit: So it turns out there is a thing called Risch's Algorithm that, properly adapted, should be able to tell me whether or not there is an antiderivative using W, or at least that is the claim in Corless, Gonnet, Hare, Jeffrey and Knuth "On the Lambert W function", p13 (search for "lambert w.pdf" (no quotes) and that paper should be the first link).
Spoiler:
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### Re: Integral of x^x

Another "breakthrough"

The integral of (1+x)x is equal to kxx*hypergeom([1, -x], [2], -x/k)

or equivalently

(kx((x+k)/k)xx+k(1+x)*((x+k)/k)x-k(1+x))/(1+x)

So if we take the limit as k goes to zero, we should have the solution. Can anybody evaluate that limit?

edit: this also may be an instance of maple doing an integral incorrectly...
Klotz
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### Re: Integral of x^x

Klotz wrote:Another "breakthrough"

The integral of (1+x)x is equal to kxx*hypergeom([1, -x], [2], -x/k)

or equivalently

(kx((x+k)/k)xx+k(1+x)*((x+k)/k)x-k(1+x))/(1+x)

So if we take the limit as k goes to zero, we should have the solution. Can anybody evaluate that limit?

edit: this also may be an instance of maple doing an integral incorrectly...

That quite neat. Before it causes too much confusion, though, it should be pointed out that you meant the integral of (k + x)x, not (1+x)x.

As for taking the limit, it's not difficult at all (just slightly messy, so forgive me for not wanting to type it out). You get an answer of (xx+1)/(x+1).

Unfortunately, this isn't right, as the derivative of that function is:

xx + (xx+1) * (ln(x) - (1/(x+1))) / (x+1), which is not quite xx. Either there's some reason we can't take the limit in this situation (most likely scenario, and I'm sure someone will tell me why this is the case) or that long formula up there involving x's and k's is wrong.

EDIT: Actually, how did you derive that formula for the hypergeometric function that you wrote up there? I'm not seeing how you got that, and it's quite possible that the problem lies there.
NathanielJ
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### Re: Integral of x^x

I simply went into Maple and wrote int( (x+k)^x,x) which gave me the hypergeometric, and then simplified to get the second term.

However, I think the integral is incorrect. The Wolfram integrator gives the integral equivalent of a 404, and if you differentiate the expression you don't get the same thing back.

What's kind of interesting is that when do an extremely naive integral or derivative, it works out. If you treat the expression you gave as a power function and take the x+1 down and reduce the exponent to x, you get x^x back.
Klotz
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### Re: Integral of x^x

Ok. If you use that integral to calculate the sophomore's dream, you get a value of 1/2, which is clearly wrong. So I don't think that integral is right.
Klotz
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### Re: Integral of x^x

Not that anyone cares, but I informed Maple of the error and they said they're working on it.
Klotz
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