R is the reals. f : R → R is continuous and L = ∫-∞∞ f(x) dx exists. Show that ∫-∞∞ f(x - 1/x) dx = L.
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
14 Jan 2002