How Fermat integrated xt dx
Theorem. (Fermat) For all rational t
other than -1, and any positive a,b, the integral of
xt dx from
x=a to x=b equals
(bt+1-at+1)/(t+1).
Proof:
Without loss of generality assume b>a.
Suppose b=rna for some integer n
and positive r. Use the partition
{xi=ria | 0<=i<=n}
of [a,b]. Then a Riemann sum for the integral is
(x1-x0)
xt0
+
(x2-x1)
xt1
+ ... +
(xn-xn-1)
xtn-1 .
This is the sum of a geometric series with initial term
(r-1)at+1,
common ratio rt+1,
and post-final term
(r-1)bt+1.
Hence the Riemann sum for this partition is
((r-1) / (rt+1-1))
·
(bt+1-at+1).
As n ``approaches infinity'', r approaches 1,
so the mesh (1-r-1)b of our partition
approaches zero. Therefore the Riemann sum approaches the integral.
So, it remains to prove that as r approaches 1
the factor ((r-1) / (rt+1-1))
tends to 1/(t+1). This is true for all nonzero t+1,
but it's particularly easy to prove when t is rational.
Let t+1=c/d, and let z be
the dth root of r.
Then we want the limit as z approaches 1 of
(zd-1)/(zc-1).
Simply remove the common factor z-1
from both numerator and denominator of this fraction;
then the numerator approaches d,
and the denominator approaches c.
Hence the limit is d/c=1/(t+1) as claimed.
QED
P.S. For irrational t we can now integrate
xt dx over [a,b]
by writing xt
as the uniform limit over that interval of xt'
with t' ranging over a sequence of rational numbers
that approach t.