From: Ronald Bruck <bruck@math.usc.edu>
Subject: Re: A Series & sone Integrals...
Date: Mon, 19 Nov 2001 19:55:41 -0800
Newsgroups: alt.math,alt.math.recreational,sci.math
Summary: Definite integral of x^(-x)

In article <UB2K7.45630$XJ4.26824002@news1.sttln1.wa.home.com>, Jeremy
Price <drjeremyphd@home.com> wrote:

> I was (believe it or not ;)) doing math for fun the other day, and I came
> upon these:
> 
> Int[ x^-x dx ] from 0 to infinity
> I don't know how to evaluate this myself, but my TI89 tells me it's about 2.
> Could anyone show me how to get a symbolic answer?

x^-x doesn't have a primitive which can be explicitly written in terms
of elementary functions; and I don't know what your integral is;
nevertheless, if you change the upper limit of integration to ONE, it
does have an explicit (and rather amazing) expansion,

     \int_0^1 x^-x dx = \sum_{n=1}^\infty 1/n^n,

which converges VERY rapidly (24 terms suffice for 32-digit precision),
to about 1.29128599706266354.

To see this, expand x^-x = exp(-x ln x) into a series,

    x^-x = \sum_{n=0}^\infty (-x log x)^n/n!

and integrate term-by-term (which must be justified).

To integrate (-x log x)^n/n! on (0,1), make the change of variable

          x = e^-t;

the value x = 0 corresponds to t = infinity, while x = 1 corresponds to
t = 0; the integral becomes

      \int_0^1 (-x log x)^n dx = \int_{\infty}^0 (t e^-t)^n *(-e^-t)dt
                               = \int_0^\infty t^n e^{-(n+1)t} dt.

Now make the change of variable (n+1)t = s, so

                               = \int_0^\infty s^n e^-s ds/(n+1)^(n+1).

But it is well-known that \int_0^\infty s^n e^-s ds = n! (this is one
of the possible definitions of the gamma function), and the desired
series falls out.

This can also be done without the gamma function.  Find a recurrence for

   \int x^m (log x)^n dx

and then take m = n.

--Ron Bruck