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Tricky (impossible?) Integral 
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#1
Jan2308, 09:02 PM

P: 119

Anyone know how to do this integral? I can't figure it out analytically, nor do I find it in my tables. Thanks!
[tex] \int x^{(x1)}dx [/tex] 


#2
Jan2408, 07:43 AM

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P: 2,616

I used the Integrator http://integrals.wolfram.com/index.jsp and got the following, which implies no elementary solutions exist:



#3
Jan2408, 07:45 AM

Math
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PF Gold
P: 39,490

Do you have any reason to believe that such a formula does exist? "Almost all" such functions do not have an antiderivative in any simple form.



#4
Jan2408, 10:27 AM

P: 119

Tricky (impossible?) Integral



#5
Jan2408, 12:17 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,490

Ah, well, if you don't require it in "simple form", I can definitely say that
[tex]\int x^{x1} dx= Ivy(x)+ C[/itex] where I have defined "Ivy(x)" to be the antiderivative of [tex]\int x^{x1}dx[/tex] such that Ivy(0)= 0! 


#6
Jan2408, 12:25 PM

P: 119




#7
Jan2408, 12:41 PM

P: 1,705

do you the antiderivative or the definite integral?



#8
Jan2408, 01:10 PM

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PF Gold
P: 12,016




#9
Jan2408, 01:20 PM

P: 119




#10
Jan2408, 02:03 PM

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PF Gold
P: 12,016

The integrand can be shown to be integrable, that's all you need to check. 


#11
Jan2408, 02:25 PM

P: 119




#12
Jan2408, 05:03 PM

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#13
Jan2408, 06:50 PM

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P: 3,348

HallsofIvy's solution is the best you will get, and theres actually nothing wrong with it either.
Here I quote Courant from page 242, Volume 1; After stating that in the 19th century it was proven certain elementary integrands did not have elementary antiderivatives 


#14
Jan2408, 06:58 PM

P: 1,705




#15
Jan2408, 07:02 PM

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P: 3,348

O ice's post reminds me; if your a desperate little one, less than a year ago I remember a paper of arVix or however you spell that site, its well known, about using hyper geometric series in a method of "approximate solutions for antiderivatives". I am not sure about the credibility of the paper, and when I skimmed through it, it seemed very elaborated. Obviously it didn't get much attention from the mathematical community, as it didn't really have any real use. Which once again brings us back to the fact that Hall's solution really actually is the best you can do in this case.
If its a definite integral, nothings stopping you from getting any degree of accuracy you want using numerical methods. 


#16
Jan2408, 07:38 PM

P: 119




#17
Jan2408, 07:47 PM

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P: 3,348

Ok. Lets go back to the days when the natural logarithm was not yet defined, and some poor souls ran into [tex]\int^x_1 \frac{1}{t} dt[/tex].
Yes, most just complained that it didn't have any nice anti derivative at all! How could they cope? But some of the bright ones said, "well nothings going to stop me from defining a new function, and showing it has these nice properties, like f(ab) = f(a) + f(b), etc etc. Ooh, hold on a second, these properties are the same ones the inverse of the exponential function must have! Hooray!" Point is, you don't need a nice analytical solution to everything to work out somethings properties and actually still achieve a "solution". It may be a nice exercise for you to prove [tex]\int^{ab}_1 \frac{1}{t} dt = \int^b_1 \frac{1}{t} dt + \int^a_1 \frac{1}{t} dt[/tex]. =] 


#18
Jan2408, 08:52 PM

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P: 2,616

Here's what Wikipedia says:



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