The integral is from -infinity to +infinity of -1/(1+x^2) dx. I know that the integral of the function in arccotx but I need help in finding what the range of values arccot has when approaching infinity. Thanks.
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By symmetry, the integral is equal to
integral of [0, infinity] -2/(1+x^2) dx
= -2 tan^-1(x) from 0 to infinity
= -pi
integral of [0, infinity] -2/(1+x^2) dx
= -2 tan^-1(x) from 0 to infinity
= -pi
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2 cot^-1(x) from 0 to infinity = 2(0 - pi/2) = -pi, the same answer.
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