2007-10-26 05:10:13 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You could use Taylor series about x = 1.
x^x =
1 + 1*(x-1) + 1*(x-1)^2 + 1/2*(x-1)^3 + 1/3*(x-1)^4 + O((x-1)^5)
Integral =
1*(x-1) + 1/2*(x-1)^2 + 1/3*(x-1)^3 + 1/8*(x-1)^4 + 1/15*(x-1)^5 + O((x-1)^6)
*EDIT*
There are many mathematical software packages available e.g. mathematica and maple. 99 out of 100 times, if these software are unable to find a closed form solution, a closed form solution does not exist.
2007-10-26 07:24:26 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
It's not "impossible". What is impossible is to
express the answer in terms of elementary functions.
In other words the integral is nonelementary.
I've often wondered: Suppose we admit
W(x) to our list of allowable functions.
This is Lambert's W function, the inverse
of f(x) = xe^x.
Can this integral be expressed in terms of W(x)
and elementary functions?
2007-10-26 06:18:08 · answer #2 · answered by steiner1745 7 · 3⤊ 0⤋
If you mean the indefinite integral of x^x, that is, its primitive, then it can't be done in terms of elementary functions. But x^x is continuous for x >0, so its definite integral exists over every compact interval [a, b] of the real line. But, using elementary functions, you can't determine such integral as F(b) - F(a), where F is a function such that F'(x) = x^x.
2007-10-26 07:00:51 · answer #3 · answered by Steiner 7 · 0⤊ 0⤋
I think so.
was that the one I saw the proofs for?
crud..
this is going to bug me.
2007-10-26 05:17:58 · answer #4 · answered by Darkwolf 5 · 0⤊ 0⤋
x^x / log x
2007-10-26 05:14:24 · answer #5 · answered by gauravragtah 4 · 0⤊ 5⤋