Evaluate ∫0∞ (tan-1(πx) - tan-1x) / x dx.
Solution
Answer: π/2 ln π.
∫0∞ (tan-1(πx) - tan-1x) / x dx = ln x (tan-1πx - tan-1x) |0∞ - ∫0∞ ( π(1 + π2x2) - 1/(1 + x2) ) ln x dx.
Putting y = πx, ∫0∞ ( π(1 + π2x2) dx = ∫0∞ (ln y - ln π)/(1 + y2) dy. So - ∫0∞ ( π(1 + π2x2) - 1/(1 + x2) ) ln x dx = ln π ∫0∞ dy/(1 + y2) = ln π tan-1y |0∞ = π/2 ln π.
It remains to show that the first term is zero. The Taylor series for tan-1x = x + O(x2), so the value at the lower limit is the limit as x → 0 of ( (π - 1) x + O(x2) ) ln x. But x ln x → 0 as x → 0, so the lower limit gives zero.
We also have the expansion tan-1x = π/2 - 1/x + O(1/x2). So the value at the upper limit is the limit of ( 1/x (1 - 1/π) + O(1/x2) ) ln x as x → ∞, which is zero since (ln x)/x → 0.
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001