-Matt is correct that the integral is:tan^-1(x) + (1/2)ln(1 + x²)This converges, if you take the difference of the two limits:lim {tan^-1(x) + (1/2)ln(1 + x²)}x→+∞- lim {tan^-1(x) + (1/2)ln(1 + x²)}x→-∞This becomes:lim tan^-1(x)x→+∞+(1/2)ln(1 + x²)x→+∞- lim tan^-1(x) + x→-∞- lim (1/2)ln(1 + x²)x→-∞The limit for the inverse tangent is clearly defined:lim tan^-1(x) = π/2x→+∞lim tan^-1(x) = - π/2x→-∞Subtract the two and you get:πThe limit for the natural log is not defined but, because x is squared in (1/2)ln(1 + x²), the limit is the same as x approaches either positive or negative infinity. Therefore,their difference results in 0.......
Once you plug in the limits from 0 to infinity, you do see that the integral diverges.
Not to sure how to prove the Cauchy Principle though.
