29th Putnam 1968

R is the reals. f : R → R is continuous and L = ∫_-∞^∞ f(x) dx exists. Show that ∫_-∞^∞ f(x - 1/x) dx = L.

Solution

Substitute x = y - 1/y. As y increases from -∞ to 0, x increases (monotonically) from -∞ to +∞. Also dx = (1 + 1/y²) dy, so we have L = ∫_-∞⁰ f(y - 1/y) (1 + 1/y²) dy = ∫_-∞⁰ f(y - 1/y) dy + ∫_-∞⁰ f(y - 1/y) 1/y² dy.

In the last integral we may substitute z = -1/y to get ∫₀^∞ f(z - 1/z) dz or, using y instead of z, ∫₀^∞ f(y - 1/y) dy.

29th Putnam 1968

© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002