How can you calculate $\int_{-\infty}^\infty\frac{dx}{x^2+1}$ using the residue at $-i$?

Question

Instead of using traditional upper-half semicircle, I have tried to calculate this integral using lower-half and have obtained $-\pi$, that is the opposite to the actual value. Can anyone show me how to calculate it?

Answer 1

Almost certainly Robert Lee is right about your error. The residue is the result of integrating around a closed curve in the counter-clockwise direction. If you traverse the curve in the clockwise direction, you get the opposite value.

If you use the upper half-circle and travel counter-clockwise, then the part of the curve on the real line is integrated from negative to positive, and taking the limit gives $\int_{-\infty}^\infty \frac{dx}{x^2 + 1}$. But if you use the lower half-circle, you can either go counter-clockwise, in which case the portion on the real line is integrated from positive to negative, so taking the limit gives $\int_{\infty}^{-\infty} \frac{dx}{x^2 + 1}$, or you can travel clockwise, in which case you do get $\int_{-\infty}^\infty$, but it is equal to the opposite of the residue, instead of the residue itself.

Either way, there is an additional negation you missed performing.