Residue theorem : Upper and Lower halfs

Question

In the following example:$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx $$ we have poles at $z=\pm i $

Now if I choose the contour in the upper complex plane, only the pole $z=i$ and I get the result as$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx=2\pi i \sum \text{Residues}=\pi $$ But if I choose to the contour in the lower complex plane I get the result as: $$\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx=2\pi i \sum \text{Residues}=-\pi $$ How do I know which contour should I choose in a given problem. Thank You.

Answer 1

The integral is equal to $\pi$. If you want to deal only with the residues from the upper half-plane, then you shall be applying the residue theorem to closed paths like the one below. And then the answer will be $\pi$, because the winding number of such a path with respect to each of the points that it surrounds is $1$.

If, on the other hand, you want to deal only with the residues from the lower half-plane, then you shall be applying the residue theorem to closed paths like the one below. And then the answer will be $\pi$ once again, because, although the residue at $-i$ is indeed $\frac i2$, the winding number of such a path with respect to each of the points that it surrounds is $-1$, not $1$ as before.

Answer 2

In the residuum theory, the curve for the integration is to be oriented in a positive way, e.g. counterclockwise. If you use a curve in a clockwise orientation instead, this simply changes the sign of the integral: Using the lower residuum, you accordingly get -(-pi)=pi.