2nd Putnam 1939

Evaluate   ∫₁³  ( (x - 1)(3 - x) )^-1/2 dx   and   ∫₁^inf  (e^x+1 + e^3-x)^-1 dx.

Solution

(1)   π. Trivial. [Write (x - 1)(3 - x) = 1 - (x - 2)². So we have a standard sin^-1z integral.]

(2)   π/(4 e²). Trivial. [Multiply top and bottom by e^x. Change variable to y = e^x-1. We now have a standard tan^-1z integral.]

2nd Putnam 1939

© John Scholes
jscholes@kalva.demon.co.uk
4 Sep 1999