Why is it difficult to integrate x^x

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I tried doing this but could not,why is it so?
 

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  • #2
dextercioby
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It's not difficult.Nobody thought of defining a special function to account for its antiderivative.

Daniel.
 
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  • #3
Zurtex
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Because it can't be integrated in terms of elementary functions. Most functions are 'not easy' to integrate in this way.
 
  • #4
HallsofIvy
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Or, to say the same thing that Dextercioby and Zurtex said, in different words, because there is no elementary function whose derivative is xx!
 
  • #5
arildno
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A better question would be:
Why is it simple to integrate f(x)=1?
 
  • #6
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i tried doin it ,but always get to a place i cant continue.who can help integrate [(x^x)(1+LOG[X])]^2.All help will be appreciated
 
  • #7
arildno
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abia ubong said:
i tried doin it ,but always get to a place i cant continue.who can help integrate [(x^x)(1+LOG[X])]^2.All help will be appreciated
You won't get any, since an anti-derivative of x^x is inexpressible in terms of elementary functions.
 
  • #8
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when i plugged it into the integrator of mathematica it gave it back as same...i dont know why it did not do computation.
 
  • #9
Zurtex
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goldi said:
when i plugged it into the integrator of mathematica it gave it back as same...i dont know why it did not do computation.
Because as said many times previously in this thread, it can not be integrated in terms of elementary functions.
 
  • #10
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i understand that yeah...but even Mathematica couldn't post the solution in terms of complex functions or whatever high level function it ocntains...
there must be a solution to it....What it is?
 
  • #11
Zurtex
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goldi said:
i understand that yeah...but even Mathematica couldn't post the solution in terms of complex functions or whatever high level function it ocntains...
there must be a solution to it....What it is?
There are no special functions defined in general mathematics for the integral.

If you want a function that is the anti-derivative of xx then just define one and then you can study its properties.
 
  • #12
matt grime
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There is a solution it is the function F such that dF/dx is x^x. But we can't write it anymore nicely than that, and there is nothing surprising about it. Almost no functions have integrals that we can write out nicely and explicitly in some closed form. How many times must that be said in this thread? Shall we lock it now to stop yet another person having to say it?
 
  • #13
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Sorry but i am such a big fool that yours terminology is not clear to me....Last time-Has its integral ever calculated ...or as ppl are saying that it has such a function as integral that has never been defined./so is research going on over this

i have a question......how can we integrate x*Sec(x)
i have tried this question than any other question ever.....
the point is that it was asked in my 12th class and when i plug it into Integrator i could not even understand the solution...
 
  • #14
arildno
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Any particular definite integral of x^x can, of course be calculated to an arbitrary degree of accuracy by numerical techniques.
 
  • #15
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To integrate x*Sec(x) I would use integration by parts, but in this case the tabular method will work nicely.

[tex]\int x\sec{x}dx[/tex]

[tex]\int udv = uv - \int vdu[/tex]

[tex]u = x[/tex]
[tex]dv = \sec{x}dx[/tex]

That should get you started.
 
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  • #16
dextercioby
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I think you meant

[tex] dv=\sec x \ dx [/tex]

I'm not sure though...:rolleyes:

Daniel.
 
  • #17
Icebreaker
Zurtex said:
If you want a function that is the anti-derivative of xx then just define one and then you can study its properties.
Has it been done before? Any interesting properties?
 
  • #18
Pyrrhus
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Look for Liouville's Principle about integration in finite terms.
 
  • #19
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Jameson said:
To integrate x*Sec(x) I would use integration by parts, but in this case the tabular method will work nicely.

[tex]\int x\sec{x}dx[/tex]

[tex]\int udv = uv - \int vdu[/tex]

[tex]u = x[/tex]
[tex]dv = \sec{x}dx[/tex]

That should get you started.
that i would have had tried 100 times...after 1 step i am stuck and there is no way out....
 
  • #20
goldi said:
i understand that yeah...but even Mathematica couldn't post the solution in terms of complex functions or whatever high level function it ocntains...
there must be a solution to it....What it is?
Well, there's F where
[tex]F(x) = \int_a^x t^t dt [/tex]
and a can be any number greater than or equal to 0. Mathematica isn't able to find that.
 
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  • #21
matt grime
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For those who think functions are *nice* (goldi etc) and aren't happy with our answers, can I ask what sin(x) is? I mean given x=2 radians say what is sin(x)? How do you define it? How do you find it? To me the answer is sin(2). There is nothing wrong with calculus's answer in the last post using the fundamental theory of calc. It is a very good function.
 
  • #22
Zurtex
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Icebreaker said:
Has it been done before? Any interesting properties?
Attached below is a picture between x = 0 and x = 3 of the function:

[tex]f(x) = \lim_{a \rightarrow 0} \int_a^x t^t dt[/tex]
 

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  • #23
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goldi said:
that i would have had tried 100 times...after 1 step i am stuck and there is no way out....
i checked into mathematica,,,it contains some functions of the form polylog but i think this is calculated with the knowledge of Complex Analysis.
 
  • #24
krab
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latyph said:
I tried doing this but could not,why is it so?
You could do it by Taylor expansion:

[tex]\int x^x dx=x + \frac{\left( -1 +
2\,\log (x) \right) \,
x^2}{4} +
\frac{\left( 2 -
6\,\log (x) +
9\,{\log (x)}^2 \right)
\,x^3}{54} +
\frac{\left( -3 +
12\,\log (x) -
24\,{\log (x)}^2 +
32\,{\log (x)}^3
\right) \,x^4}{768} +
\frac{\left( 24 -
120\,\log (x) +
300\,{\log (x)}^2 -
500\,{\log (x)}^3 +
625\,{\log (x)}^4
\right) \,x^5}{75000} +
\frac{\left( -5 +
30\,\log (x) -
90\,{\log (x)}^2 +
180\,{\log (x)}^3 -
270\,{\log (x)}^4 +
324\,{\log (x)}^5
\right) \,x^6}{233280}
+ {O(x^7)[/tex]
 
  • #25
shmoe
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I wonder if students aren't done a disservice in first year calculus classes with their sterilized examples and problems. They'll be asked to do hundreds of integration problems, all rigged to work out nicely with the techniques they've just learned. Perhaps it will be mentioned that there are functions whose antiderivatives cannot be written in a "nice" form, but examples will be scarce- [tex]e^{-x^2}[/tex] being the stock one. After seeing such an unnatural ratio of nice examples to possibly one or two 'not-nice' ones, it's no suprise that many walk away believing themselves invincible and any function that itself looks 'nice' will have a 'nice' antiderivative waiting around the corner so they flap their arms around and bash their heads in frustration trying to find it. Makes me wonder if they bother to even consider why numerical techniques are taught at all?
 

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