integration of x^x
Ernesto Gianola
Can anyone tell me how to go about solving this integral
Integral[ x^x dx]
I've had the toughest time with it and have gotten nowhere.
JOHN WIMMER
>From: ne...@bu.edu (Ernesto Gianola)
>Subject: integration of x^x
>Date: 29 Mar 93 15:24:04 GMT
Matthew P Wiener
>Can anyone tell me how to go about solving this integral
>Integral[ x^x dx]
It cannot be integrated in elementary terms. Ray Steiner recently reposted
an old article of mine, giving the differential algebraic proof of a basic
theorem of Liouville on integration in elementary terms. Moreover, I
followed up his question about applying this theorem to (arcsin x)/x.
The following may seem overly abbreviated if you haven't seen the above
articles. I'll e-mail copies of them to you if you wish, but if what
follows looks too jargonish to you as opposed to merely too sketchy,
you probably won't find them of much use.
------------------------------------------------------------------------
MAIN THEOREM. Let F,G be differential fields, let a be in F, let y be in G,
and suppose y'=a and G is an elementary differential extension field of F,
and Con(F)=Con(G). Then there exist c_1,...,c_n in Con(F), u_1,...,u_n, v
in F such that
u_1' u_n'
a = c_1 --- + ... + c_n --- + v'.
u_1 u_n
In other words, the only functions that have elementary anti-derivatives
are the ones that have this very specific form.
------------------------------------------------------------------------
In this case, let F=C(z,l)(t), the field of rational functions in z,l,t,
where l=log z and t=exp(zl)=z^z. Note that z,l,t are algebraically
independent. (Choose some appropriate domain of definition.) Then
t'=(1+l)t, so for a=t in the above situation, the partial fraction
analysis (of the sort done in the previous posts) shows that the only
possibility is for v=wt+... to be the source of the t term on the left,
with w in C(z,l).
So this means, equating t coefficients, 1=w'+(l+1)w. This is a first
order ODE, whose solution is w=I(z^z)/z^z. So we must prove that no
such w exists in C(z,l). So suppose (as in one of Ray Steiner's posts)
w=P/Q, with P,Q in C[z,l] and having no common factors. Then z^z=
(z^z*P/Q)'=z^z*[(1+l)PQ+P'Q-PQ']/Q^2, or Q^2=(1+l)PQ+P'Q-PQ'. So Q|Q',
meaning Q is a constant, which we may assume to be one. So we have
it down to P'+P+lP=1.
Let P=Sum[P_i l^i], with P_i, i=0...n in C[z]. But then in our equation,
there's a dangling P_n l^(n+1) term, a contradiction.
--
-Matthew P Wiener (wee...@sagi.wistar.upenn.edu)
E. Johnson
>
>Integral[ x^x dx]
Well, isn't it just [x^(x+1)]/(x+1)?
Because according to the power rule of differentiation,
if f(x) = x^n, f'(x) = nx^(n-1)
then
int [nx^(n-1)] dx = x^n
and
int [x^n] dx = [x^(n+1)]/(n+1).
So wouldn't that be your solution? (If you already know all about the
integration power rule, sorry.)
-eaj
Robert E George
writes:
>In article <113...@bu.edu> ne...@bu.edu (Ernesto Gianola) writes:
>>
>>Can anyone tell me how to go about solving this integral
>>
>>Integral[ x^x dx]
>
>Well, isn't it just [x^(x+1)]/(x+1)?
No, it's not.
>
>Because according to the power rule of differentiation,
>
> if f(x) = x^n, f'(x) = nx^(n-1)
>
>then
>
> int [nx^(n-1)] dx = x^n
>
>and
>
> int [x^n] dx = [x^(n+1)]/(n+1).
>
Correct statement of the power rule. However . . .
>So wouldn't that be your solution? (If you already know all about the
>integration power rule, sorry.)
The power rule as you describe it only applies if the exponent is a
constant -- when the exponent is itself a variable, then you cannot
use the power rule.
>
>-eaj
>
Robert George
(speaking only for myself)
Michael L. Harrison
>Can anyone tell me how to go about solving this integral
>Integral[ x^x dx]
This is not complete, but one approach which simplifies the
situation is write:
x = Exp[Log[x]]
Thus, x^x = Exp[x Log[x]]
Here, the well known techniques may be of some use.
-MH
Erik Max Francis
> In article <113...@bu.edu> ne...@bu.edu (Ernesto Gianola) writes:
> >
> >Can anyone tell me how to go about solving this integral
> >
> >Integral[ x^x dx]
>
> Well, isn't it just [x^(x+1)]/(x+1)?
>
> Because according to the power rule of differentiation,
The power rule deals with a _constant_ exponent. x is not a constant,
and so the power rule can't apply. After all, try differentiating x^x.
It's rather a different beastie than x^n.
Erik Max Francis, &tSftDotIotE ...!apple!uuwest!max m...@west.darkside.com __
USMail: 1070 Oakmont Dr. #1 San Jose, CA 95117 ICBM: 37 20 N 121 53 W / \
If you like strategic games of interstellar conquest, ask about UNIVERSE! \__/
-)(- Omnia quia sunt, lumina sunt. All things that are, are lights. -)(-
peter dunn
>ne...@bu.edu (Ernesto Gianola) writes:
>>Integral[ x^x dx]
>-MH
In a similar way, let y = x^x.
Then, ln(y) = x ln(x).
Implicit diff. gives:
(1/y) (dy/dx) = 1 + ln(x)
so (dy/dx) = y(1+ln(x))
x
= x (1 + ln(x))
Well, at least, I think this is OK...?
Have fun... enjoy...
--
\ ~~~~~This~posting~is~brought~to~you~by:~Peter~Dunn...~Enjoy...~~~ /
--\-- Faculty of Science, USQ : e-mail: du...@zeus.usq.edu.au--/--
\ Toowoomba, QLD, AUSTRALIA 4350 : Phone: (076) 31 277 /
\___"War is not about who is right; it's about who is left."____/
Timothy Kimball
: harr...@ugcs.caltech.edu (Michael L. Harrison) writes:
: >ne...@bu.edu (Ernesto Gianola) writes:
: >>Can anyone tell me how to go about solving this integral
: >>Integral[ x^x dx]
: >This is not complete, but one approach which simplifies the
: >situation is write:
: >x = Exp[Log[x]]
: >Thus, x^x = Exp[x Log[x]]
: >Here, the well known techniques may be of some use.
: >-MH
: In a similar way, let y = x^x.
: Then, ln(y) = x ln(x).
: Implicit diff. gives:
: (1/y) (dy/dx) = 1 + ln(x)
: so (dy/dx) = y(1+ln(x))
: x
: = x (1 + ln(x))
: Well, at least, I think this is OK...?
It's fine. It's not the _integral_ of x^x, though.
--
/* tdk -- Science Computing and Research Support (SCARS) -- STScI */
/* Disclaimer: The above remarks are my My MY personal observations. */
Ray Steiner
>Integral[ x^x dx]
Rewrite it as int(exp(xln(x)). A try on MAPLE also yielded
no results. I'm almost sure the integral is nonelementary.
I will try to use the ideas of my last few posts(and those
of M. Wiener) to prove this. More to follow!
Ray Steiner
--
ste...@andy.bgsu.edu
Erik Max Francis
> so (dy/dx) = y(1+ln(x))
> x
> = x (1 + ln(x))
>
> Well, at least, I think this is OK...?
Yep, looks pretty good to me. Now try (d/dx) x^x^x -- that is,
(d/dx) x^(x^x).
(d/dx) (x^x)^x is pretty trivial, as that's the same thing as (d/dx) x^(2
x).
Soeren Kjaer Soerensen
In article <1993Mar31....@stsci.edu> kim...@stsci.edu (Timothy Kimball) writes:
[stuff deletet]
: >ne...@bu.edu (Ernesto Gianola) writes:
: >>Can anyone tell me how to go about solving this integral
: >>Integral[ x^x dx]
: >This is not complete, but one approach which simplifies the
: >situation is write:
: >x = Exp[Log[x]]
: >Thus, x^x = Exp[x Log[x]]
: >Here, the well known techniques may be of some use.
: >-MH
: In a similar way, let y = x^x.
: Then, ln(y) = x ln(x).
: Implicit diff. gives:
: (1/y) (dy/dx) = 1 + ln(x)
: so (dy/dx) = y(1+ln(x))
: x
: = x (1 + ln(x))
: Well, at least, I think this is OK...?
Kimball> It's fine. It's not the _integral_ of x^x, though.
Yeps, your right. But it shows another point, ei you can integrate the sum
x^x(1 + ln(x)) but NOT the terms x^x and x^x ln(x).
S0REN.
--
/ | From this proposition it will follow, when
_//( | arithmetical addition has been defined, that 1+1=2.
/ | \ |
s...@iesd.auc.dk | Whitehead & Russel, p. 360 PRINCIPIA MATHEMATICA