Evaluate ∫_{0}^{∞} (tan^{-1}(πx) - tan^{-1}x) / x dx.

**Solution**

Answer: π/2 ln π.

∫_{0}^{∞} (tan^{-1}(πx) - tan^{-1}x) / x dx = ln x (tan^{-1}πx - tan^{-1}x) |_{0}^{∞} - ∫_{0}^{∞} ( π(1 + π^{2}x^{2}) - 1/(1 + x^{2}) ) ln x dx.

Putting y = πx, ∫_{0}^{∞} ( π(1 + π^{2}x^{2}) dx = ∫_{0}^{∞} (ln y - ln π)/(1 + y^{2}) dy. So - ∫_{0}^{∞} ( π(1 + π^{2}x^{2}) - 1/(1 + x^{2}) ) ln x dx = ln π ∫_{0}^{∞} dy/(1 + y^{2}) = ln π tan^{-1}y |_{0}^{∞} = π/2 ln π.

It remains to show that the first term is zero. The Taylor series for tan^{-1}x = x + O(x^{2}), so the value at the lower limit is the limit as x → 0 of ( (π - 1) x + O(x^{2}) ) ln x. But x ln x → 0 as x → 0, so the lower limit gives zero.

We also have the expansion tan^{-1}x = π/2 - 1/x + O(1/x^{2}). So the value at the upper limit is the limit of ( 1/x (1 - 1/π) + O(1/x^{2}) ) ln x as x → ∞, which is zero since (ln x)/x → 0.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001