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/y^{2}) dy, so we have L = ∫_{-∞}^{0} f(y - 1/y) (1 + 1/y^{2}) dy = ∫_{-∞}^{0} f(y - 1/y) dy + ∫_{-∞}^{0} f(y - 1/y) 1/y^{2} 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.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002