What is your shortest proof of the following integral identity?

Question

⌠¹
⌡ x^(ax) dx = 1 - a/(2^2) + (a^2)/(3^3) - (a^3)/(4^4) + ···
º
(I'll settle for proofs where a=0,1,2,3..., if necessary)

This text editor doesnt do math problems justice. The integral is from 0 to 1 of the function x^(ax).
The sum is Σ (-a)^k/(k+1)^(k+1) ; k = 0, 1, 2, 3, ... ∞

Mark Taranto
This is essentially the proof that I have now. It is rather involved to be rigorous however. I was wondering if anyone had one a little more elegant...

Answer 1

First let I(k,n)=int (log x)^k x^n dx where the integral goes from 0 to 1. Do an integration by parts with u=(log x)^k x^(n-1) and dv=x dx
to get that I(k,n)=-k/(n+1). By induction, we then obtain that
I(n,n)=(-1)^n n!/(n+1)^n I(0,n)
=(-1)^n n!/(n+1)^(n+1).

Now, write x^(ax)=exp(ax log x)=
sum 1/n! (ax log x)^n=
sum (1/n!) a^n x^n (log x)^n.

Now integrate. Since the sum converges uniformly, the interchange of the sum and integral gives
int x^(ax) dx=sum (1/n!) a^n int x^n (log x)^n dx
= sum (1/n!) a^n I(n,n)
=sum (-1)^n a^n /(n+1)^(n+1)
as desired.

Answer 2

Try substituting the Taylor Series expansion for x^(ax). You will need to refer to some dominated convergence theorem to show that the integral of the infinite sum is equal to the infinite sume of the integral -- but this should do the trick.

Answer 3

I'll look at this after the football match, hope some people get a chance at it.

Yeah, what mathematician said; too much beer during football match . . .