30th Putnam 1969

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Problem A4

Prove that ∫01 xx dx = 1 - 1/22 + 1/33 - 1/44 + ... .

 

Solution

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
jscholes@kalva.demon.co.uk
14 Jan 2002