Prove that ∫01 xx dx = 1 - 1/22 + 1/33 - 1/44 + ... .
The rhs is a series, so this suggests that we should expand the integrand as a series and integrate term by term.
We have that xx = ex ln x, so the obvious approach is to expand this as a series: 1 + (x ln x) + (x ln x)2/2! + ... . For this to work we need that ∫01 (x ln x)n dx = n! (-1)n/(n+1)n+1 (*).
The integral is easy to evaluate by parts. Each step reduces the exponent on the log term without affecting the exponent on the x term and the non-integral term always vanishes at both endpoints. In other words, we have ∫01 xn lnmx dx = -m/(n+1) ∫01 xn lnm-1x dx, which gives (*).
30th Putnam 1969
© John Scholes
14 Jan 2002