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intx^x

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Hi there.

I'm having problems integrating [math]y=x^x[/math]

Any help would be appreciated.

I was going to say:

 

(x^x+1)/(x+1) + C

x cannot = -1

 

won't this work??

Whilst it's certainly true that [math]\int x^n \ dx = \frac{x^{n+1}}{n+1} + c[/math] (assming n not equal to -1), you can't do this for things like xx - we're assuming that the n in the first equation is constant.

  • Author

Why is it not intergratable though?

If you drew it, it would have a meaningful area between x=0 and x=5? would it not?

I do remember a thread on x^x a few weeks back, but i don't remember what was said exactly.

 

Cheers guys.

Is there a reason why it won't integrate or is it just a 'rule'? (if you know what i mean)

  • Author

If i remember rightly, it was that it doesn't have any calculable area or something like that.

If you actually try to integrate it:

[math]y=x^x[/math]

[math]logy=xlogx[/math]

therefore y=e^(xlogx)

If we used integration by parts...

[math]integrate 1.e^(xlogx)[/math]

u=e^(xlogx) dv=1

It just gets more and more complicated...

ie du=x^x(logx+1)dx; v=x

[math]x.e^(xlogx)-integrate e^(xlogx)(logx+1)dx[/math]

Thats not going anywhere!

so x^(x+1)-int x^x.(LNx+1) just as bad....

 

sorry for the poor use of LaTeX, as i'm just figuring it out

I think it is integrable, but cannot be expressed with elementary functions, so the result will inevitably be something like an infinite series.

it is integrable in any sense of the word, but the function that is its anti-derivative is not expressable in elementary terms. this is no surprise, alomst all functions are like this. sadly people are preconditioned to believing that the baby examples they're taught in school reflect a general pattern that holds "in the real world"

 

nb in the real world means, effectively, in a mathematical situation that arises from studyign something rather than the special cases dreamt up as exercises.

One can, however, evaluate the definite integral numberically.

 

One can also say that [imath]\int x^x dx = bling(x) [/imath], where bling is a function that is equal to this integral.

why hmm? that is perfectly true and follows the wittgenstien approach of something is its definition. what is -1? it is the number that when added to 1 gives 0. it is the correct and necessary approach to many parts of mathematics, or so some of us would say. what is sin(35.657357)? what is pi? cos most people mistakenly think decimals are real numbers, and that the answers on calculator are correct doesnj't mean they are. i once saw a text book used in schools that staetd 1.41 was the square root of two.

  • Author

I admit to my ignorance Matt. But if it "is integratable in every sense of the word" why can you not prove it? What is the definition of this function, as this seems to be the crux of the matter?

 

Thank you.

it is integrable; you know what integrable is? it does not mean having a nice antiderivative that you can write down in terms of nice functions. x^x is (away from 0) riemann integrable, lesbegue integrable and stiltjes integrable, hence integrable in every sense. the definition of its antiderivative is exactly what DQW wriote for "bling", or rather thatis one of its anti derivatives, bling +k for any real k would also do.

I should probably add that you can integrate pretty much every function numerically, so even when it's not possible analytically, you can still obtain fairly good estimates.

 

 

And that is true of all functions for almost all inputs. it is slightly disigenuous to pretend that sin is somehow distinctly better than "bling" as above.

[sidetrack]If Dave's a 'boing', can I be a 'bling' ? [/sidetrack]

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